RS Aggarwal Class 9 Mathematics Solutions Chapter 11 Circle

Exercise 11A

Question 1. Let AB be a chord of the given circle with centre O and radius 10 cm. Then, OA = 10 cm and AB = 16 cm. From O, draw OL ⊥ AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord.
Answer: Since the perpendicular from the centre meets a chord and cuts it in half, we find AL = (1/2) × AB = (1/2) × 16 = 8 cm. Using the right-angled triangle OLA, we apply the Pythagorean theorem: OA² = OL² + AL². Substituting the values: 10² = OL² + 8², which gives us 100 = OL² + 64. Therefore, OL² = 36, so OL = 6 cm. The distance of the chord from the centre is 6 cm.
In simple words: When you drop a line from the centre of the circle straight down to the chord at a right angle, it splits the chord into two equal pieces. Using the Pythagorean theorem with the radius and half the chord length, we can find how far away the chord sits from the centre.

Exam Tip: Always recognise that a perpendicular from the centre to a chord creates a right-angled triangle - use the Pythagorean theorem immediately.

Question 2. Let AB be the chord of the given circle with centre O and radius 5 cm. From O, draw OL ⊥ AB. Then, OA = 5 cm and OL = 3 cm (given). We know that the perpendicular from the centre of a circle to a chord bisects the chord.
Answer: By using the perpendicular bisector property of circles, the line from the centre that meets a chord at right angles divides the chord into two equal halves. In right-angled triangle OLA, we use the Pythagorean theorem: OA² = AL² + OL². Rearranging: AL² = OA² - OL² = 5² - 3² = 25 - 9 = 16. Therefore, AL = 4 cm. Since the perpendicular bisects the chord, AB = 2 × AL = 2 × 4 = 8 cm. The length of the chord is 8 cm.
In simple words: The perpendicular from the centre splits the chord evenly. Using the Pythagorean theorem with the radius and the distance from centre to chord, you get half the chord length, then multiply by 2 to find the full chord.

Exam Tip: Remember the perpendicular bisector property - it creates two congruent right triangles, making calculations straightforward.

Question 4. Let AB be the chord of the given circle with centre O. Draw OL ⊥ AB.
Answer: We are given that AB = 30 cm and OL = 8 cm, where OL represents the distance from the centre to the chord. Using the perpendicular bisector property, AL = (1/2) × AB = (1/2) × 30 = 15 cm. In right-angled triangle OLA, we apply the Pythagorean theorem: OA² = OL² + AL² = 8² + 15² = 64 + 225 = 289. Therefore, OA = √289 = 17 cm. The radius of the circle is 17 cm. For a second pair of chords where CD = 6 cm and MB = NC = 8 cm and OB = OD = 5 cm (radius), we find LB = (1/2) × AB = 4 cm and MD = (1/2) × CD = 3 cm. In right-angled triangle BLO, we get LO² = OB² - LB² = 5² - 4² = 25 - 16 = 9, so LO = 3 cm. In right-angled triangle DMO, we get MO² = OD² - MD² = 5² - 3² = 25 - 9 = 16, so MO = 4 cm. The distance between the chords is (4 - 3) = 1 cm.
In simple words: Find how far each chord sits from the centre using the Pythagorean theorem. The difference between these two distances gives you how far apart the chords are.

Exam Tip: For distance between parallel chords, calculate the perpendicular distance from centre to each chord separately, then subtract to get the gap between them.

Question 5. (ii) Let AB and CD be two chords of a circle such that AB || CD and they are on the opposite sides of the centre. AB = 8 cm and CD = 6 cm. Draw OL ⊥ AB and OM ⊥ CD.
Answer: Join OA and OC to identify the radius. Then OA = OC = 5 cm (radius). Since the perpendicular from the centre to a chord bisects it, AL = (1/2) × AB = (1/2) × 8 = 4 cm. Similarly, CM = (1/2) × CD = (1/2) × 6 = 3 cm. In right-angled triangle OLA, we find OL² = OA² - AL² = 5² - 4² = 25 - 16 = 9, so OL = 3 cm. In right-angled triangle OMC, we find OM² = OC² - CM² = 5² - 3² = 25 - 9 = 16, so OM = 4 cm. Since the chords are on opposite sides of the centre, the distance between them is OL + OM = 3 + 4 = 7 cm.
In simple words: When two chords lie on opposite sides of the centre, find the perpendicular distance from the centre to each chord. Add these distances together to get the total gap between the chords.

Exam Tip: Pay careful attention to whether chords are on the same side or opposite sides of the centre - this determines whether you add or subtract the perpendicular distances.

Question 6. Let AB and CD be two chords of a circle having centre O. AB = 30 cm and CD = 16 cm.
Answer: Join AO and OC, which are both radii. Draw OM ⊥ CD and OL ⊥ AB. Using the property that the perpendicular from the centre to a chord bisects the chord, AL = (1/2) × AB = (1/2) × 30 = 15 cm and CM = (1/2) × CD = (1/2) × 16 = 8 cm. In right-angled triangle ALO, we have AO² = OL² + AL² → OL² = AO² - AL² = 17² - 15² = 289 - 225 = 64, so OL = 8 cm. In right-angled triangle CMO, we have OC² = OM² + CM² → OM² = OC² - CM² = 17² - 8² = 289 - 64 = 225, so OM = 15 cm. The distance between the chords is OM + OL = 15 + 8 = 23 cm.
In simple words: Calculate the perpendicular distance from the centre to each chord using the Pythagorean theorem. Add both distances to find how far the chords sit from each other.

Exam Tip: Always apply the perpendicular bisector theorem first, then use the Pythagorean theorem systematically for each chord.

Question 7. CD is the diameter of a circle with centre O, and is perpendicular to chord AB. Join CA.
Answer: Let OA = OC = r cm. Then OE = (r - 3) cm, where CE = 3 cm. We are given AB = 12 cm and CE = 3 cm. Since the perpendicular from the centre to a chord bisects the chord, AE = (1/2) × AB = (1/2) × 12 = 6 cm. In right-angled triangle CEA, we apply the Pythagorean theorem: OA² = OE² + AE² → r² = (r - 3)² + 6². Expanding: r² = r² - 6r + 9 + 36 → 0 = -6r + 45 → 6r = 45 → r = 7.5 cm. Therefore, OA, the radius of the circle is 7.5 cm.
In simple words: The perpendicular from the centre bisects the chord into two equal parts. Setting up the Pythagorean equation using the radius, the distance from centre to the given point on the diameter, and half the chord, you can solve for the radius.

Exam Tip: When a diameter is perpendicular to a chord, the bisected chord and the diameter create a right triangle - use this to find the radius algebraically.

Question 8. AB is the diameter of a circle with centre O which bisects the chord CD at point E. CF = ED = 8 cm and EB = 4 cm. Join OC. Let OC = OB = r cm. Then OE = (r - 4) cm.
Answer: We are given that CD is the diameter bisecting perpendicular at E, with CF = ED = 8 cm and EB = 4 cm. Using the perpendicular bisector property, when AB (the diameter) is perpendicular to CD at E, the right-angled triangle OEC forms: OC² = OE² + EC² → r² = (r - 4)² + 8². Expanding: r² = r² - 8r + 16 + 64 → 0 = -8r + 80 → 8r = 80 → r = 10 cm. The radius of the circle is 10 cm.
In simple words: When the diameter stands perpendicular to another chord, it cuts that chord in half. Use the Pythagorean relationship with the radius, the offset distance, and half the chord to find the radius value.

Exam Tip: Recognise the perpendicular bisector setup immediately - it always creates a right triangle where the radius is the hypotenuse.

Question 9. Given: OD ⊥ AB of a circle with centre O. BC is a diameter. To Prove: AC || OD and AC = 2 × OD. Construction: Join AC.
Answer: We know that the perpendicular from the centre of a circle to a chord bisects the chord. Since OD meets AB perpendicularly, D is the midpoint of AB, so AD = DB. Also, O is the midpoint of BC (as BC is a diameter). By the Midpoint Theorem, the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equals half of it. In triangle ABC, D is the midpoint of AB and O is the midpoint of BC, therefore OD || AC and OD = (1/2) × AC, which means AC = 2 × OD.
In simple words: The perpendicular from the centre splits the chord in half. When combined with the diameter property, the Midpoint Theorem tells us that a line connecting two midpoints runs parallel to the third side and measures exactly half of it.

Exam Tip: Spot the Midpoint Theorem application - identify the two midpoints clearly before stating the parallel and proportional relationships.

Question 10. Given: O is the centre in which chords AB and CD intersect at P such that PO bisects ∠BPD. To Prove: AB = CD. Construction: Draw OE ⊥ AB and OF ⊥ CD.
Answer: In triangles OEP and OFP, we have ∠OEP = ∠OFP (each equal to 90°), OP = OP (common side), and ∠OPE = ∠OPF (since OP bisects ∠BPD). By the Angle-Side-Angle criterion of congruence, triangles OEP and OFP are congruent. Therefore, the corresponding parts are equal: OE = OF. This means both chords sit at the same perpendicular distance from the centre. Equal chords of a circle are equidistant from the centre, so AB = CD.
In simple words: When you draw perpendiculars from the centre to both chords and show they have equal length, then the chords themselves must be equal. The perpendicular distances from the centre determine chord equality.

Exam Tip: Use congruent triangles to prove that perpendicular distances are equal, then apply the equidistant chord theorem to finish the proof.

Question 11. If possible let two different circles intersect at three distinct point A, B and C. Then, these points are noncollinear. So a unique circle can be drawn to pass through these points. This is a contradiction.
Answer: Suppose two different circles both pass through three non-collinear points A, B, and C. By the fundamental principle of circles, exactly one unique circle can be drawn through any three non-collinear points. Since both circles pass through all three points, they must be identical circles, contradicting our assumption that they are different. Therefore, two distinct circles cannot intersect at three different points - they can intersect at most at two points.
In simple words: Three non-collinear points determine exactly one circle, no more and no less. If two different circles both passed through three such points, they would have to be the same circle, which is impossible.

Exam Tip: Use proof by contradiction - assume the opposite and show it leads to an impossibility based on the uniqueness of the circle through three points.

Question 12. OA = 10 cm and AB = 12 cm.
Answer: We have two circles with centres O and O'. Since the perpendicular from the centre to a chord bisects the chord, AD = (1/2) × AB = (1/2) × 12 = 6 cm. In right-angled triangle ADO, we find OD² = OA² - AD² = 10² - 6² = 100 - 36 = 64, so OD = 8 cm. Similarly, for the second circle with centre O', O'A = 8 cm. In right-angled triangle ADO', we find O'D² = O'A² - AD² = 8² - 6² = 64 - 36 = 28, so O'D = √28 = 2√7 cm. The distance between the circle's centres is OO' = (OD + O'D) = (8 + 2√7) cm.
In simple words: The perpendicular from each centre splits the shared chord equally. Find each centre's distance to the chord using the Pythagorean theorem, then add them to get the distance between the two centres.

Exam Tip: When two circles intersect, the line joining their centres is perpendicular to the common chord - use this and the Pythagorean theorem for each circle.

Question 13. Given: Two equal circles intersect at points P and Q. A straight line through P meets the circles in A and B. To Prove: QA = QB. Construction: Join PQ.
Answer: Two circles will be congruent if and only if they have equal radii. If two chords of a circle are equal then their corresponding arcs are congruent. Here PQ is the common chord to both the circles. Thus, their corresponding arcs are equal. So, arc PCQ = arc PDQ → ∠QAP = ∠QBP (congruent arcs have the same degree measure). In an isosceles triangle, base angles are equal. Therefore, QA = QB.
In simple words: Equal circles share a common chord PQ. This chord subtends equal arcs in both circles. As a result, the angles at A and B (where a line through P meets the circles) are equal, making the triangle isosceles with QA = QB.

Exam Tip: Congruent circles have equal radii and thus equal corresponding arcs - use this to establish angle equalities and then apply isosceles triangle properties.

Question 14. Given: AB and CD are the two chords of a circle with centre O. Diameter POQ bisects them at L and M. To Prove: AB || CD.
Answer: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at L and M. Then, OL ⊥ AB and OM ⊥ CD. Therefore, ∠ALM = ∠CMD (alternate angles are equal). This means AB || CD (alternate angles are equal).
In simple words: When a diameter is perpendicular to two different chords, it makes equal angles with both. These equal angles tell us the chords must be parallel to each other.

Exam Tip: When perpendiculars from a point meet two lines, if they make equal angles, the lines are parallel - use alternate angle properties.

Question 15. Two circles with centres A and B, having radii 5 cm and 3 cm touch each other internally. The perpendicular bisector of AB meets the bigger circle in P and Q. Join AP.
Answer: Since the circles touch each other internally, their centres A and B and the points of tangency are collinear. Let PQ intersect AB at L. Then AB = (5 - 3) = 2 cm. Since PQ is the perpendicular bisector of AB, we have AL = (1/2) × AB = (1/2) × 2 = 1 cm. In right-angled triangle PLA, we find PL² = AP² - AL² = 5² - 1² = 25 - 1 = 24 cm² → PL = √24 = 2√6 cm. Since PQ = (2 × PL) = (2 × 2√6) = 4√6 cm, the length of PQ is 4√6 cm.
In simple words: The perpendicular bisector of the line joining two circle centres cuts that line exactly in half. Using the Pythagorean theorem with the larger radius and half the distance between centres, you find each half of the perpendicular, then double it for the full length.

Exam Tip: The perpendicular bisector of a line segment creates right triangles with any point on itself - exploit this to apply the Pythagorean theorem.

Question 17. Given: AB is a chord of a circle with centre O. AB is produced to C such that BC = OB. Also, CO is joined to meet the circle in D. ∠ACD = y° and ∠AOD = x°.
Answer: Since OB = BC, triangle OBC is isosceles. Therefore, ∠BOC = ∠BCO = y°. Extended, ∠OBA = ∠BOC + ∠BCO = (2y)°. Since OA = OB (both radii), triangle OAB is isosceles, so ∠OAB = ∠OBA = (2y)°. Exterior angle ∠AOD = ∠OAC + ∠ACO = (2y)° + y° = 3y° → x° = 3y°.
In simple words: Since OB equals BC, the triangle OBC has two equal sides, making two angles equal. This leads to other angle relationships through isosceles triangles, ultimately showing that the angle at the centre is three times the angle at the exterior point.

Exam Tip: Identify isosceles triangles formed by radii or equal segments - base angles in isosceles triangles are always equal, simplifying angle chasing.

Question 18. Given: AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC.
Answer: Since AB = AC (given), and equal chords of a circle are equidistant from the centre, we have OM = ON (where M and N are feet of perpendiculars from O to the chords). The perpendicular from the centre to a chord bisects the chord. Therefore, MB = NC. Since OP ⊥ AB and OQ ⊥ AC, and the perpendiculars from the centre make equal distances from the centre, considering triangles OMB and ONc, we can establish that PB = QC by the properties of equal chords. Thus, by Side-Angle-Side criterion of congruence, we have triangle MPB ≅ triangle NQC. The corresponding parts of the congruent triangles are equal, so PB = QC.
In simple words: Equal chords always sit at the same distance from the centre. When perpendiculars are drawn from the centre to equal chords, they create congruent geometric figures, meaning segments cut off on the chords are equal.

Exam Tip: Remember that equal chords are equidistant from the centre - this property cascades into other equalities via triangle congruence.

Question 19. Let triangle ABC be an equilateral triangle of side 9 cm. Let AD be one of its medians. Then, AD ⊥ BC and BD = (1/2) × BC = (1/2) × 9 = 4.5 cm.
Answer: In an equilateral triangle with side 9 cm, the median from A to the midpoint D of BC is perpendicular to BC. Using the Pythagorean theorem in the right triangle ADB: AD² = AB² - BD² = 9² - (4.5)² = 81 - 20.25 = 60.75 → AD = √60.75 = (9√3)/2 cm. In an equilateral triangle, the centroid and circumcentre coincide, and AG:GD = 2:1. Therefore, the radius AG = (2/3) × AD = (2/3) × (9√3)/2 = 3√3 cm. The radius of the circumscribed circle is 3√3 cm.
In simple words: An equilateral triangle's median is also its altitude. The circumcentre is located at the centroid, dividing the median in a 2:1 ratio from the vertex. This ratio directly gives the circumradius.

Exam Tip: For an equilateral triangle, the circumradius equals (side length)/√3 or (2/3) of the median height - memorise this shortcut.

Question 20. Given: BC is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD. To Prove: AB = CD. Construction: Draw OL ⊥ AB and OM ⊥ CD.
Answer: Since OL ⊥ AB and OM ⊥ CD, and AB || CD, the perpendiculars from O to these parallel chords are also perpendicular to each other in a specific geometric arrangement. Using the perpendicular bisector theorem, OL bisects AB and OM bisects CD. In right-angled triangles OLB and OMD, since ∠OLB = ∠OMD = 90° (perpendicular bisector, angle = 90°), and given that AB || CD and BC is a diameter (making OB = OC = radius), we establish that triangles OLB and OMD are congruent by Angle-Angle-Side. Therefore, OL = OM and LB = MD, which means the full chords satisfy AB = CD.
In simple words: Parallel chords that both relate to the same diameter create congruent triangles when perpendiculars from the centre are drawn. Equal perpendicular distances and symmetric geometry ensure the chords themselves are equal.

Exam Tip: When chords are parallel and the circle has a specific diameter involved, use the symmetric properties created by that diameter to establish congruence.

Question 21. Given: AB and AC are two equal chords of a circle with centre O. To Prove: ∠OAB = ∠OAC. Construction: Join OA, OB and OC.
Answer: In triangles OAB and OAC, we have AB = AC (given equal chords), OA = OA (common side), and OB = OC (both radii). By the Side-Side-Side criterion of congruence, triangle OAB ≅ triangle OAC. The corresponding parts of congruent triangles are equal, therefore ∠OAB = ∠OAC. This also means that O lies on the bisector of ∠BAC, so OA bisects the angle between the two equal chords.
In simple words: Two equal chords sharing a common endpoint create two triangles with the centre. Since all three sides match (the equal chords, the common point, and the equal radii), the triangles are identical, making the angles at the common point equal.

Exam Tip: When two chords are equal and meet at a point, immediately consider the two triangles formed with the centre - their congruence follows from SSS.

Exercise 11B

Question 1. (i) Join BO.
Answer: In triangle BOC, OC = OB (both equal to the radius). Therefore, angle OBC = angle OCB (base angles of an isosceles triangle are equal). Since angle OCB = 30°, we have angle OBC = 30°. Thus, angle OBC = angle OCB. By the angle sum property, angle BOC + angle OBC + angle OCB = 180° → angle BOC + 30° + 30° = 180° → angle BOC = 120°. Now, in triangle OAB, OA = OB (both radii). Therefore, angle OAB = angle OBA (base angles are equal). Since we determined angle OBA = 40° (given), angle OAB = 40°. By angle sum, angle AOB + 40° + 40° = 180° → angle AOB = 100°. The angle subtended by an arc at the centre is double the angle subtended at any point on the circumference. Thus, angle AOC = 2 × angle ABC → 100° + 120° = 140° (from parts 1 and 2). We know angle BOC = 2 × angle BAC → 120° = 2 × angle BAC → angle BAC = 60°. Therefore, angle ABC = 70°.
In simple words: Equal sides in a triangle mean equal angles opposite those sides (isosceles triangle property). Using angle sums and the inscribed angle theorem, which states the central angle is twice the inscribed angle, you can find all required angles systematically.

Exam Tip: Always identify isosceles triangles formed by radii, and remember that the central angle subtended by an arc is exactly double the inscribed angle subtended by the same arc.

Question 2. (i) The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
Answer: Given angle AOB = 70° at the centre, the angle ACB subtended by the same arc AB at a point C on the circumference is half the central angle. Therefore, angle ACB = (1/2) × angle AOB = (1/2) × 70° = 35°. (ii) The radius of the circle is OA = OC. Since OA = OC (both radii), triangle OAC is isosceles, so the base angles are equal: angle OAC = angle OCA (base angles of isosceles triangle are equal). By angle sum, angle OAC = 35° (as angle OCA = 35°).
In simple words: The angle at the centre is always twice any angle at the edge that subtends the same arc. This is the inscribed angle theorem - it works for any point on the circle.

Exam Tip: Use the inscribed angle theorem immediately when you see a central angle and an inscribed angle subtending the same arc - one is always double the other.

Question 4. [Diagram and partial content shown - incomplete question text]
Answer: It is clear that BD is the diameter of the circle. We also know that the angle in a semicircle is a right angle, so angle BAD = 90°. Now consider triangle BAD. Using the angle sum property: angle ADB + angle DAB + angle ABD = 180° → angle ADB + 90° + 35° = 180° → angle ADB = 55°. Angles in the same segment of a circle are equal, so angle ACB = angle ADB = 55°. Therefore, angle ACB = 55°.
In simple words: Any angle inscribed in a semicircle (subtended by the diameter) is a right angle. Once you identify the diameter, you immediately know one angle is 90°, then use angle sum to find others. Angles in the same circular segment are identical.

Exam Tip: The angle in a semicircle theorem is powerful - recognise the diameter and instantly mark that angle as 90°, then proceed with basic angle arithmetic.

Question 6. (i) Angles in the same segment of a circle are equal. ∠ABD and ∠ACD are in the segment AD. Therefore, ∠ACD = ∠ABD = 54° (given). (ii) Angles in the same segment of a circle are equal. ∠BAD and ∠BCD are in the segment BD. Therefore, ∠BAD = ∠BCD = 43° (given). (iii) Consider the triangle ABD. By angle sum property we have: ∠BAD + ∠ADB + ∠DBA = 180° → 43° + ∠ADB + 54° = 180° → ∠ADB = 180° - 97° = 83°. Angles in the same segment of a circle are equal. ∠BDA and ∠BCA are in the segment BA. Therefore, ∠BDA = 83°.
Answer: Angles sitting in the same circular segment - that is, angles subtended by the same arc at different points on the circumference - are always equal. This is why ∠ABD = ∠ACD = 54° (both subtend arc AD). Similarly, ∠BAD = ∠BCD = 43° (both subtend arc BD). In triangle ABD, the three angles must sum to 180°. Given ∠BAD = 43° and ∠DBA = 54°, we calculate ∠ADB = 180° - 43° - 54° = 83°. Since ∠BDA and ∠BCA both subtend the same arc BA from different circumference points, they too are equal, so ∠BCA = 83°.
In simple words: All angles drawn from different points on a circle to the same arc are identical. Use this fact to identify equal angles, then apply the angle sum property in triangles to find unknown angles.

Exam Tip: Angles in the same segment are your best friend - spot them immediately and mark multiple angles as equal without calculation.

Question 8. Angles in the same segment of a circle are equal. \( \angle CAD \) and \( \angle CBD \) are in the segment CD. \( \angle CAD = \angle CBD = 60° \). We know that an angle in a semi circle is a right angle. \( \angle ADC = 90° \) [angle in a semicircle]. \( \angle ACD = 180° - (\angle ADC + \angle CAD) = 180° - (90° + 60°) = 180° - 150° = 30° \). \( \angle CDE = \angle ACD = 30° \) [AC | DE and CD is a transversal, thus alternate angles are equal].
Answer: The angles located in the same segment of a circle will always be equal. Since \( \angle CAD \) and \( \angle CBD \) both lie in segment CD, they are equal at \( 60° \). In a semicircle, any inscribed angle measures \( 90° \). Therefore, \( \angle ADC = 90° \). Using the angle sum property in triangle ACD, we get \( \angle ACD = 180° - 90° - 60° = 30° \). When AC is parallel to DE and CD acts as a transversal, the alternate interior angles are equal, so \( \angle CDE = 30° \).
In simple words: Angles sitting in the same curved section of a circle are always the same size. If one of them is 60 degrees, the other is also 60 degrees. This rule helps us figure out angles in circles.

Exam Tip: Always identify angles in the same segment first - they are automatically equal. Use the "angle in a semicircle is 90°" property to find missing angles quickly.

Question 9. (i) \( \angle CED = 90° \). In \( \triangle CED \), we have \( \angle CED + \angle EDC + \angle DCE = 180° \). \( 90° + 40° + \angle DCE = 180° \). \( \angle DCE = 180° - 130° = 50° \) ...(1). (ii) \( \angle AOC \) and \( \angle BOC \) are linear pair. \( \angle BOC = (180° - 80°) = 100° \) ......(2). \( \angle ABC = 180° - (\angle BOC + \angle DCE) = 180° - (100° + 50°) = 180° - 150° = 30° \).
Answer: (i) In triangle CED, the angle at E is a right angle. The three interior angles must total 180°, so \( 90° + 40° + \angle DCE = 180° \), which gives \( \angle DCE = 50° \). (ii) The angles AOC and BOC form a linear pair, making them supplementary. Therefore, \( \angle BOC = 180° - 80° = 100° \). Finally, \( \angle ABC \) can be found by: \( \angle ABC = 180° - (100° + 50°) = 30° \).
In simple words: First, find the missing angle in the triangle by making sure all three angles add up to 180 degrees. Then, use the fact that angles on a straight line always total 180 degrees. Combine these results to get your final answer.

Exam Tip: Remember that angles on a straight line sum to 180° and angles in any triangle also sum to 180°. Use these two properties together to solve multi-step problems.

Question 11. The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference. \( \angle AOB = 2\angle ACB \). \( = 2\angle DCB \) [\( \angle ACB = \angle DCB \)]. \( \angle DCB = \frac{1}{2}\angle AOB = \left(\frac{1}{2} \times 40\right) = 20° \). Consider the \( \triangle BEC \); By angle sum property, we have \( \angle BDC + \angle DCB + \angle DBC = 180° \). \( 100° + 20° + \angle DBC = 180° \). \( \angle DBC = 180° - 120° = 60° \). \( \angle OBC = \angle DBC = 60° \). \( \angle OBC = 60° \).
Answer: An arc of a circle always subtends an angle at the centre that is twice the angle it subtends at any point on the circumference. Since \( \angle ACB = \angle DCB \), we have \( \angle AOB = 2\angle DCB \). Substituting the given value: \( \angle DCB = \frac{1}{2} \times 40° = 20° \). In triangle BEC, applying the angle sum property: \( 100° + 20° + \angle DBC = 180° \), so \( \angle DBC = 60° \). Therefore, \( \angle OBC = 60° \).
In simple words: The angle at the centre of a circle is always twice as big as the angle at the edge (the circumference). If the central angle is 40 degrees, the angle at the edge is 20 degrees. Then use the triangle angle rule to find the remaining angle.

Exam Tip: The "angle at centre is double the angle at circumference" rule is crucial - apply it immediately when you see a central angle and a circumference angle subtending the same arc.

Question 12. Join OB. OA = OB [Radius]. \( \angle OBA = \angle OAB = 25° \) [base angles are equal in isosceles triangle]. Now in \( \triangle OAB \), we have \( \angle OAB + \angle OBA + \angle AOB = 180° \). \( 25° + 25° + \angle AOB = 180° \). \( \angle AOB = 180° - 50° = 130° \). The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference. \( \angle AOB = 2\angle ACB \). \( \angle ACB = \frac{1}{2}\angle AOB = \frac{1}{2} \times 130° = 65° \). \( \angle ECB = 65° \).
Answer: By joining point O to B, we establish that OA and OB are both radii, making them equal. In an isosceles triangle, the base angles are equal, so \( \angle OBA = \angle OAB = 25° \). Using the angle sum property in triangle OAB: \( 25° + 25° + \angle AOB = 180° \), giving us \( \angle AOB = 130° \). The central angle theorem tells us that the central angle is twice the inscribed angle subtending the same arc. Therefore, \( \angle ACB = \frac{1}{2} \times 130° = 65° \), which means \( \angle ECB = 65° \).
In simple words: Join the centre O to point B. Since both OA and OB are radii from the circle's centre, they must be equal in length. This makes triangle OAB isosceles, so the two base angles are each 25 degrees. Add them up with the central angle and you get the angle at the circumference using the doubling rule.

Exam Tip: When you see a radius mentioned, immediately think "isosceles triangle" - this opens up the use of equal base angles. Combine this with the central angle theorem for a quick solution.

Exercise 11C

Question 1. (i) \( \angle BDC = \angle BAC = 40° \) [angles in the same segment]. In \( \triangle BCD \), we have \( \angle BCD + \angle BDC + \angle DBC = 180° \). \( \angle BCD + 40° + 60° = 180° \). \( \angle BCD = 180° - 100° = 80° \). (ii) Also, \( \angle CAD = \angle CBD \) [angles in the same segment]. \( \angle CAD = 60° \) [\( \angle CBD = 60° \)].
Answer: (i) Angles located in the same segment of a circle are equal, so \( \angle BDC = \angle BAC = 40° \). In triangle BCD, the sum of all interior angles equals 180°, giving us: \( \angle BCD + 40° + 60° = 180° \), which means \( \angle BCD = 80° \). (ii) Similarly, angles in the same segment satisfy \( \angle CAD = \angle CBD \). Since \( \angle CBD = 60° \), we have \( \angle CAD = 60° \).
In simple words: If two angles sit in the same curved section of a circle (the same segment), they must be equal. Use this rule along with the triangle angle sum to find all unknown angles.

Exam Tip: Always spot angles in the same segment first - they save you calculation steps. Then apply the triangle angle sum property to complete the solution.

Question 3. In cyclic quadrilateral PQRS, \( \angle PSR + \angle PQR = 180° \). \( 15° + \angle PQR = 180° \). \( \angle PQR = 180° - 15° = 30° \) ......(i). Also, \( \angle PRQ = 90° \) ......(ii) [angle in a semi circle].
Answer: In a cyclic quadrilateral, the property states that opposite angles are supplementary (sum to 180°). Therefore, \( \angle PSR + \angle PQR = 180° \), which gives us \( 15° + \angle PQR = 180° \), so \( \angle PQR = 30° \). Additionally, since \( \angle PRQ \) is inscribed in a semicircle, it must equal \( 90° \).
In simple words: When a quadrilateral sits inside a circle, its opposite angles always add up to exactly 180 degrees. That's a key rule for cyclic quadrilaterals. Also, any angle drawn inside a semicircle is always a right angle.

Exam Tip: Cyclic quadrilateral means opposite angles sum to 180° - use this as your first property. Semicircle angles are always 90° - remember this special case.

Question 4. In cyclic quadrilateral PQRS, \( \angle PSR + \angle PQR = 180° \). \( 15° + \angle PQR = 180° \). \( \angle PQR = 180° - 15° = 30° \) ......(i). Also, \( \angle PRQ = 90° \) ......(ii) [angle in a semi circle]. Now in \( \triangle PRQ \) we have \( \angle PQR + \angle FRQ + \angle RPQ = 180° \). \( 30° + 90° + \angle RPQ = 180° \) [from (i) and (ii)]. \( \angle RPQ = 180° - 120° = 60° \).
Answer: In the cyclic quadrilateral PQRS, opposite angles sum to 180°. So \( \angle PSR + \angle PQR = 180° \), giving us \( \angle PQR = 30° \). Since \( \angle PRQ \) sits in a semicircle, it equals \( 90° \). In triangle PRQ, the angle sum property gives: \( 30° + 90° + \angle RPQ = 180° \), so \( \angle RPQ = 60° \).
In simple words: Use the cyclic quadrilateral rule first to find one angle. Then use the semicircle angle rule. Finally, apply the triangle angle sum rule to find the last missing angle.

Exam Tip: Layer your properties: start with the cyclic quadrilateral property, then apply the semicircle property, and finish with triangle angles. This step-by-step approach minimises errors.

Question 6. ABCD is a cyclic quadrilateral. \( \angle ABC + \angle ADC = 180° \). \( 92° + \angle ADC = 180° \). \( \angle ADC = 180° - 92° = 88° \). Also, AE | CD. \( \angle EAD = \angle ADC = 88° \). \( \angle BCD = \angle DAF \) [- exterior angle of a cyclic quadrilateral = int opp angle]. \( \angle BCD = \angle EAD + \angle EAF = 88° + 20° = 108° \) [\( \angle FAE = 20°(given) \)]. \( \angle BCD = 108° \).
Answer: Since ABCD is a cyclic quadrilateral, opposite angles are supplementary. Therefore, \( \angle ABC + \angle ADC = 180° \), which gives us \( \angle ADC = 88° \). Since AE is parallel to CD, alternate interior angles are equal, so \( \angle EAD = 88° \). An exterior angle of a cyclic quadrilateral equals the opposite interior angle, so \( \angle BCD = \angle EAD + \angle EAF = 88° + 20° = 108° \).
In simple words: In a cyclic quadrilateral, opposite angles add up to 180 degrees. When you have a parallel line, use alternate angles being equal. Combine these two ideas to find the exterior angle.

Exam Tip: When parallel lines appear in a cyclic quadrilateral problem, alternate interior angles become your shortcut. Look for them immediately.

Question 8. Angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment. Here arc ABC makes \( \angle AOC = 100° \) at the centre of the circle and \( \angle ADC \) on the circumference of the circle. \( \angle AOC = 2\angle ADC \). \( \angle ADC = \frac{1}{2}(\angle AOC) \). \( = \frac{1}{2} \times 100° \) [\( \angle AOC = 100° \)]. \( \angle ADC = 50° \). The opposite angles of a cyclic quadrilateral are supplementary, ABCD is a cyclic quadrilateral and thus, \( \angle ADC + \angle ABC = 180° \) [- \( \angle CDE = 50° \)] and \( \angle ABC = 130° \). \( \angle ADC = 50° \) and \( \angle ABC = 130° \).
Answer: An arc subtends an angle at the centre that is double the angle it subtends at any point on the circumference in the alternate segment. Arc ABC creates a central angle of \( 100° \) and a circumference angle \( \angle ADC \). Using the relationship: \( \angle ADC = \frac{1}{2} \times 100° = 50° \). In cyclic quadrilateral ABCD, opposite angles must be supplementary, so \( \angle ADC + \angle ABC = 180° \). Therefore, \( \angle ABC = 180° - 50° = 130° \).
In simple words: The central angle is always twice as large as the angle at the circle's edge (circumference). Once you have one angle in a cyclic quadrilateral, the angle opposite to it must add up to 180 degrees with it.

Exam Tip: Central angles are always double the circumference angles for the same arc. Then use the opposite angles property in cyclic quadrilaterals to find all remaining angles quickly.

Question 9. ABCD is a cyclic quadrilateral. \( \angle A + \angle C = 180° \) [opp.angle of a cyclic quadrilateral are supplementary]. \( \angle A + 100° = 180° \). \( \angle A = 180° - 100° = 80° \). Now in \( \triangle ABD \), we have \( \angle A + \angle ABD + \angle ADB = 180° \). \( 80° + 50° + \angle ADB = 180° \). \( \angle ADB = 180° - 130° = 50° \). \( \angle ADB = 50° \).
Answer: In a cyclic quadrilateral, opposite angles are supplementary. Therefore, \( \angle A + \angle C = 180° \), which gives \( \angle A = 80° \). Using the angle sum property in triangle ABD: \( 80° + 50° + \angle ADB = 180° \), so \( \angle ADB = 50° \).
In simple words: Opposite angles in a cyclic quadrilateral always add up to 180 degrees. Once you know one angle, you can find its opposite. Then use triangle angle sums for the remaining parts.

Exam Tip: The supplementary property of opposite angles in cyclic quadrilaterals is your starting point - use it every time to find one angle quickly.

Question 11. O is the centre of the circle and \( \angle BOD = 150° \). Reflex \( \angle BOD = (360° - \angle BOD) = (360° - 150°) = 210° \). Now, \( x = \frac{1}{2}(\text{reflex } \angle BOD) = \frac{1}{2} \times 210° = 105° \). \( x = 105° \). Again, \( x + y = 180° \). \( 105° + y = 180° \). \( y = 180° - 105° = 75° \). \( y = 75° \).
Answer: The reflex angle BOD is obtained by subtracting the given angle from 360°: Reflex \( \angle BOD = 360° - 150° = 210° \). An angle at the circumference is half the central angle, so \( x = \frac{1}{2} \times 210° = 105° \). Since angles x and y form a linear pair, they are supplementary: \( x + y = 180° \), giving us \( y = 180° - 105° = 75° \).
In simple words: A reflex angle is the larger angle going the "long way around" a circle - it's 360 degrees minus the smaller angle. The angle at the circle's edge is half the reflex central angle. Angles on a straight line add up to 180 degrees.

Exam Tip: When you see "reflex angle", immediately calculate 360° minus the given angle. Then apply the central angle rule using the reflex angle, not the standard angle.

Question 12. ABCD is a cyclic quadrilateral. We know that in a cyclic quadrilateral exterior angle = int erior opposite angle. \( \angle CBF = \angle CDA = (180° - x) \). \( 130° = 180° - x \). \( x = 180° - 130° = 50° \). \( x = 50° \).
Answer: For a cyclic quadrilateral, an exterior angle equals the interior opposite angle. Therefore, \( \angle CBF = \angle CDA \), which means \( 130° = 180° - x \), so \( x = 50° \).
In simple words: When a side of a cyclic quadrilateral is extended outward to make an exterior angle, that exterior angle always equals the angle at the opposite vertex inside the quadrilateral.

Exam Tip: The exterior angle property for cyclic quadrilaterals is a powerful shortcut - use it whenever a side is extended beyond the quadrilateral.

Question 14. AB and CD are two chords of a circle which intersect each other at P, outside the circle. AB = 6cm, BP = 2 cm and PD = 2.5 cm. Therefore, AP x BP = CP x DP. Or, 8 x 2 = (CD + 2.5) x 2.5 cm [as CP = CD + DP]. Let x = CD. Thus, \( 8 \times 2 = (x + 2.5) \times 2.5 \). \( 16 \text{ cm} = 2.5 x + 6.25 \text{ cm} \). \( 2.5x = (16 - 6.25) \text{ cm} \). \( 2.5x = 9.75 \text{ cm} \). \( x = \frac{9.75}{2.5} = 3.9 \text{ cm} \). \( x = 3.9 \text{ cm} \). Therefore, CD = 3.9 cm.
Answer: When two chords intersect outside a circle at point P, the power of a point theorem states: AP \( \times \) BP = CP \( \times \) DP. With AB = 6 cm and BP = 2 cm, we get AP = 8 cm. Substituting: \( 8 \times 2 = (CD + 2.5) \times 2.5 \). Letting x = CD: \( 16 = 2.5x + 6.25 \), so \( 2.5x = 9.75 \), giving us \( x = 3.9 \) cm. Therefore, CD = 3.9 cm.
In simple words: When two chords (or their extensions) cross at a point outside the circle, multiply the two distances from that point to the circle along one line, and this product equals the product of the two distances along the other line. This lets you solve for any unknown length.

Exam Tip: The power of a point theorem is essential for intersecting chords/secants. Always set up the equation as product on one line = product on the other line, then solve carefully with algebra.

Question 16. Consider the triangles, \( \triangle EBC \) and \( \triangle EDA \). Side AB of the cyclic quadrilateral ABCD is produced to E. \( \angle EBC = \angle CDA \). \( \angle EBC = \angle EDA \). ......(i). Again, side DC of the cyclic quadrilateral ABCD is produced to E. \( \angle ECB = \angle BAD \). \( \angle ECB = \angle EAD \). .......(ii). and \( \angle BEC = \angle DEA \) [each equal to \( \angle E \)] .......(iii). Thus from (i), (ii) and (iii), we have \( \triangle EBC \cong \triangle EDA \).
Answer: Consider triangles EBC and EDA. When side AB is extended to point E, the exterior angle equals the interior opposite angle in the cyclic quadrilateral: \( \angle EBC = \angle CDA = \angle EDA \). Similarly, when DC is extended to E: \( \angle ECB = \angle BAD = \angle EAD \). Both triangles share angle E, so \( \angle BEC = \angle DEA \). By the Angle-Angle-Angle (AAA) similarity criterion, or by having all corresponding angles equal, we conclude \( \triangle EBC \cong \triangle EDA \).
In simple words: When you extend the sides of a cyclic quadrilateral and look at the two triangles formed outside, their angles match up perfectly. The exterior angles of the quadrilateral match the interior opposite angles, and both triangles share the same angle at point E.

Exam Tip: The exterior angle property of cyclic quadrilaterals creates angle equalities that allow you to prove triangle congruence. Always extend sides and check for equal angles in the resulting triangles.

Question 17. \( \triangle ABC \) is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E. Since AB = AC. \( \angle ACB = \angle ABC \). So, ext. \( \angle ADE = \angle ACB = \angle ABC \). \( \angle ADE = \angle ABC \). \( \Rightarrow \) DE || BC.
Answer: In the isosceles triangle ABC where AB = AC, the base angles are equal: \( \angle ACB = \angle ABC \). The circle through B, C, D, and E creates a cyclic quadrilateral, so the exterior angle at D equals the interior opposite angle: \( \angle ADE = \angle ACB \). Since \( \angle ADE = \angle ABC \) and these are corresponding angles, we conclude DE || BC.
In simple words: In an isosceles triangle, the two base angles are always equal. A circle passing through two vertices and intersecting two sides creates angle relationships that force the chord inside to be parallel to the base.

Exam Tip: Isosceles triangles combined with circles are powerful - the equal base angles and the cyclic quadrilateral property work together to create parallel lines. Spot this pattern early.

Question 19. \( \triangle ABC \) is an isosceles triangle in which AB = AC. D and E are the mid points of AB and AC respectively. DE || BC. \( \angle ADE = \angle ABC \) ......(i). Also, AB = AC [Given]. \( \angle ABC = \angle ACB \) ......(ii). \( \angle ADE = \angle ACB \) [From (i) and (ii)]. Now, \( \angle ADE + \angle EDB = 180° \) [- ADis a straight line]. \( \angle ACB + \angle EDB = 180° \). \( \Rightarrow \) The opposite angles are supplementary. \( \Rightarrow \) D, B, C and E are concyclic. ie, D, B, C and E is a cyclic quadrilateral.
Answer: In isosceles triangle ABC with AB = AC, the base angles are equal: \( \angle ABC = \angle ACB \). Since D and E are midpoints, DE is parallel to BC. For a line parallel to BC, the angle it makes with AB equals the base angle: \( \angle ADE = \angle ABC \). Therefore, \( \angle ADE = \angle ACB \). Since AD is a straight line, \( \angle ADE + \angle EDC = 180° \), which means \( \angle ACB + \angle CDB = 180° \). Because these opposite angles sum to 180°, the quadrilateral DBCE is cyclic, making D, B, C, and E concyclic.
In simple words: Start with an isosceles triangle and join the midpoints - this creates a parallel line. The angles this parallel line makes match the base angles of the triangle. When opposite angles in a quadrilateral add to 180 degrees, all four points must lie on a circle.

Exam Tip: Midpoints and parallel lines in isosceles triangles hint at cyclic quadrilaterals. Test whether opposite angles sum to 180° to confirm all four points are concyclic.

Question 20. ABCD is a rhombus. Let the diagonals AC and BD of the rhombus ABCD intersect at O. But, we know, that the diagonals of a rhombus bisect each other at right angles. So, \( \angle BOC = 90° \). \( \angle BOC \) lies in a circle. Thus the circle drawn with BC as diameter will pass through O. Similarly, all the circles described with AB, AD and CD as diameters will pass through O.
Answer: In a rhombus ABCD, the diagonals AC and BD intersect at point O. A key property of rhombuses is that the diagonals bisect each other at right angles, so \( \angle BOC = 90° \). Since angle BOC is a right angle, it lies on the circle with BC as diameter. By the same reasoning, circles drawn with AB, AD, and CD as diameters will all pass through O, as the angle subtended by each side at O is 90°.
In simple words: A rhombus has diagonals that cross at right angles (90 degrees) at their midpoint. Any angle of 90 degrees automatically lies on a circle whose diameter is the opposite side. So all four circles formed this way pass through the intersection point of the diagonals.

Exam Tip: The right angle property of rhombus diagonals is key - it immediately creates four 90° angles at the centre. Use the "angle in a semicircle is 90°" rule in reverse to identify circles.

Question 22. Let A, B, C be the given points. With B as centre and radius equal to AC draw an arc. With C as centre and AB as radius draw another arc, which cuts the previous arc at D. Then D is the required point BD and CD. In \( \triangle ABC \) and \( \triangle DCB \). AB = DC. AC = DB. BC = CB [common]. \( \triangle ABC \cong \triangle DCB \) [by SSS]. \( \angle BAC = \angle CDB \) [CPCT]. Thus, BC subtends equal angles, \( \angle BAC \) and \( \angle CDB \) on the same side of it. \( \therefore \) Points A, B, C, D are concyclic.
Answer: To locate point D, draw an arc centred at B with radius AC, and another arc centred at C with radius AB. These arcs intersect at D. In triangles ABC and DCB, we have AB = DC (by construction), AC = DB (by construction), and BC is common. By the Side-Side-Side (SSS) criterion, \( \triangle ABC \cong \triangle DCB \). Therefore, corresponding angles are equal: \( \angle BAC = \angle CDB \). Since BC subtends equal angles at points A and D on the same side, the four points A, B, C, D must be concyclic.
In simple words: Use compass and ruler to build point D so that the two triangles become congruent. Congruent triangles have equal angles. When a chord subtends equal angles at two points on the same side, those two points and the two endpoints of the chord all lie on one circle.

Exam Tip: Triangle congruence (SSS) produces equal angles, which immediately triggers the concyclic condition. Always look for equal angles subtended by the same chord on the same side.

Question 23. ABCD is a cyclic quadrilateral. \( \angle B - \angle D = 60° \) ......(i). and \( \angle B + \angle D = 180° \) ......(ii). Adding (i) and (ii) we get, \( 2\angle B = 240° \). \( \angle B = \frac{240°}{2} = 120° \). Substituting the value of \( \angle B = 120° \) in (i) we get. \( 120° - \angle D = 60° \). \( \angle D = 120° - 60° = 60° \). The smaller of the two angles i.e.\( \angle D = 60° \).
Answer: In cyclic quadrilateral ABCD, opposite angles are supplementary, so \( \angle B + \angle D = 180° \) ......(ii). We are also given that \( \angle B - \angle D = 60° \) ......(i). Adding equations (i) and (ii): \( 2\angle B = 240° \), which gives \( \angle B = 120° \). Substituting back into equation (i): \( 120° - \angle D = 60° \), so \( \angle D = 60° \). The smaller angle is \( \angle D = 60° \).
In simple words: You have two pieces of information: the angles add to 180 degrees (the cyclic quadrilateral rule) and their difference is 60 degrees (given). Solve these two equations together using algebra - add them to eliminate one variable, then substitute back.

Exam Tip: When you have both a sum and a difference of two quantities, add the equations to eliminate one variable, then substitute to find the other. This technique works for any pair of linear equations.

Question 25. ABCD is a quadrilateral in which AD = BC and \( \angle ADC = \angle BCD \). Draw DE \( \perp \) AB and CF \( \perp \) AB. Now, in \( \triangle ADE \) and \( \triangle BCF \), we have \( \angle AED = \angle BFC \) [each equal to 90°]. \( \angle ADE = \angle ADC - 90° = \angle BCD - 90° = \angle BCF \). AD = BC [given]. Thus, by Angle-Angle-Side criterion of congruence, we have \( \triangle ADE \cong \triangle BCF \) [by AAS congruence]. The corresponding parts of the congruent triangles are equal. \( \angle A = \angle B \). Now, \( \angle A + \angle B + \angle C + \angle D = 360° \). \( \angle A + \angle B + 2\angle D = 360° \) [\( \angle C = \angle D \)]. \( 2\angle B + 2\angle D = 360° \). \( 2(\angle B + \angle D) = 360° \). \( \angle B + \angle D = 180° \). \( \therefore \) ABCD is a cyclic quadrilateral.
Answer: Draw perpendiculars DE and CF from D and C to line AB respectively. In triangles ADE and BCF, both have right angles at E and F: \( \angle AED = \angle BFC = 90° \). Since \( \angle ADC = \angle BCD \) and both perpendiculars subtract 90° from these angles, we get \( \angle ADE = \angle BCF \). Given AD = BC, by the Angle-Angle-Side (AAS) criterion, \( \triangle ADE \cong \triangle BCF \). Therefore, \( \angle A = \angle B \). Using the angle sum property for quadrilaterals: \( \angle A + \angle B + \angle C + \angle D = 360° \). Since \( \angle C = \angle D \), we have \( 2\angle B + 2\angle D = 360° \), so \( \angle B + \angle D = 180° \). This means ABCD is cyclic.
In simple words: Drop perpendiculars to create two right triangles that share special properties. Use triangle congruence to prove two angles of the quadrilateral are equal. Then show that opposite angles add to 180 degrees, which is the defining property of cyclic quadrilaterals.

Exam Tip: Constructing perpendiculars in quadrilateral problems often creates congruent triangles. Once you prove congruence, corresponding angles become equal - this is a powerful tool for cyclic quadrilateral proofs.

Question 26. Given: Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles. Let OL \( \perp \) AB such that LO produced meets CD at M. To Prove: CM = MD. Proof: \( \angle 1 = \angle 2 \) [angles in the same segment]. \( \angle 2 + \angle 3 = 90° \) [\( \angle AOB = 90° \)] and \( \angle 3 + \angle 4 = 90° \) [\( \angle LOM \) is a straight line and \( \angle BOC = 90° \)]. \( \angle 2 + \angle 3 = \angle 3 + \angle 4 \). \( \angle 2 = \angle 4 \). Thus, \( \angle 1 = \angle 2 \) and \( \angle 2 = \angle 4 \). \( \angle 1 = \angle 4 \). \( \therefore \) OM = CM. Similarly, OM = MD. Hence, CM = MD.
Answer: In cyclic quadrilateral ABCD with perpendicular diagonals intersecting at O, we draw OL perpendicular to AB, and extend it to meet CD at M. We need to show CM = MD. By the angles in the same segment rule, \( \angle 1 = \angle 2 \). Since \( \angle AOB = 90° \), we have \( \angle 2 + \angle 3 = 90° \). Similarly, since \( \angle BOC = 90° \) and LOM is a straight line, \( \angle 3 + \angle 4 = 90° \). Equating these two sums: \( \angle 2 = \angle 4 \). Combined with \( \angle 1 = \angle 2 \), we get \( \angle 1 = \angle 4 \). Therefore, OM = CM. By the same reasoning, OM = MD. Hence, CM = MD.
In simple words: Perpendicular diagonals in a cyclic quadrilateral create special angle relationships. Angles in the same segment are always equal. Use these equal angles to show that two distances from M (to C and D) must be the same, making M the midpoint of CD.

Exam Tip: Perpendicular diagonals are a special property - use the 90° angles they create to build chains of equal angles. Look for isosceles triangles formed by equal angles, which give equal sides.

Question 27. Chord AB of a circle is produced to E. If one side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. Ext. \( \angle BDE = \angle BAC = \angle EAC \) ......(1). Chord CD of a circle is produced to E. Ext. \( \angle DBE = \angle ACD = \angle ACE \) ......(2). Consider the triangles \( \triangle EDB \) and \( \triangle EAC \). \( \angle BDE = \angle CAE \) [from(1)]. \( \angle DBE = \angle ACE \) [from(2)]. \( \angle E = \angle E \) [common]. \( \therefore \) \( \triangle EDB \approx \triangle EAC \).
Answer: When chord AB is extended to E, the exterior angle of cyclic quadrilateral equals the interior opposite angle: Ext. \( \angle BDE = \angle BAC = \angle EAC \) ......(1). Similarly, when chord CD is extended to E: Ext. \( \angle DBE = \angle ACD = \angle ACE \) ......(2). In triangles EDB and EAC, we have \( \angle BDE = \angle CAE \) from (1), \( \angle DBE = \angle ACE \) from (2), and both triangles share angle E. By the Angle-Angle-Angle (AAA) criterion, \( \triangle EDB \sim \triangle EAC \).
In simple words: Extend two sides of a cyclic quadrilateral to a point E outside. The exterior angles created equal certain interior angles. These equal angles appear in two triangles that share a common angle at E, proving the triangles are similar.

Exam Tip: The exterior angle property of cyclic quadrilaterals is powerful for creating similar triangles. Look for angle equalities from this property, then spot the shared angle at the point where extensions meet.

Question 28. Given: AB and CD are two parallel chords of a circle BDE and ACE are straight lines which intersect at E. If one side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. Ext. \( \angle EDC = \angle A \) and Ext. \( \angle DCE = \angle A \). \( \angle A = \angle B \). \( \therefore \) \( \triangle AEB \) is isosceles. Also, AB || CD. \( \angle EDC = \angle B \) and \( \angle DCE = \angle A \). \( \angle A = \angle B \).
Answer: Since AB and CD are parallel chords in a circle, and lines BDE and ACE are straight extensions intersecting at E, the exterior angle property of cyclic quadrilaterals applies. The exterior angle at D equals the interior angle at A: Ext. \( \angle EDC = \angle A \). Similarly, Ext. \( \angle DCE = \angle A \). Since both exterior angles equal \( \angle A \), we have \( \angle EDC = \angle DCE = \angle A \). Using the parallel chord property \( AB || CD \) and the angle equalities derived, triangle AEB becomes isosceles with \( \angle A = \angle B \).
In simple words: Two parallel chords and lines extending from their endpoints create a special configuration. The exterior angles formed by extending one chord equal interior angles. This makes certain triangle angles equal, creating an isosceles triangle.

Exam Tip: Parallel chords combined with the exterior angle property of cyclic quadrilaterals often create isosceles triangles. Identify the equal angles first, then spot the isosceles triangle.

Question 29. AB is a diameter of a circle with centre O. ADE and CBE are straight lines, meeting at E, such that \( \angle BAD = 35° \) and \( \angle BED = 25° \). Join BD and AC. (i) Now, \( \angle BDA = 90° = \angle EDB \) [angle in a semi circle]. \( \angle EBD = 180° - (\angle EDB + \angle BED) = 180° - (90° + 25°) = 180° - 115° = 65° \). \( \angle DBC = (180° - \angle EBD) = 180° - 65° = 115° \). \( \angle DBC = 115° \). (ii) Again, \( \angle DCB = \angle BAD \) [angle in the same segment]. Since, \( \angle BAD = 35° \). \( \angle DCB = 35° \). (iii) \( \angle BDC = 180° - (\angle DBC + \angle DCB) = 180° - (115° + 35°) = 180° - 150° = 30° \). \( \angle BDC = 30° \).
Answer: (i) Since AB is a diameter, any angle subtended by it at the circumference is a right angle: \( \angle BDA = 90° \). In triangle EBD, the angle sum gives: \( \angle EBD = 180° - (90° + 25°) = 65° \). The angle \( \angle DBC \) is supplementary to \( \angle EBD \) on line E - B - C: \( \angle DBC = 180° - 65° = 115° \). (ii) Angles in the same segment of a circle are equal, so \( \angle DCB = \angle BAD = 35° \). (iii) In triangle BDC, the angle sum property gives: \( \angle BDC = 180° - (115° + 35°) = 30° \).
In simple words: A diameter of a circle always creates a 90-degree angle at the circumference (angle in a semicircle). Use this rule, then triangle angle sums, then the "angles in the same segment" rule to find all angles step by step.

Exam Tip: Diameter + circumference angle = 90° is automatic - use it immediately. Then layer the triangle angle sums and the same-segment property to solve systematically.