RS Aggarwal Class 9 Mathematics Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle

Question 1. AB = AC implies their opposite angle are equal
Answer: In triangle ABC, if AB = AC, the triangle is isosceles. The angles opposite to the equal sides are equal. Since AB and AC are equal, the angles opposite to them - which are angle C and angle B - must also be equal. Therefore, ∠B = ∠C = 55°.
In simple words: When two sides of a triangle are the same length, the angles across from them are also the same size.

Exam Tip: Always identify which angles are opposite to the equal sides - this is the key to solving isosceles triangle problems quickly.

Question 2. Consider the isosceles triangle ∆ABC. Since the vertical angle of ABC is 100°, we have, ∠A = 100°.
Answer: In an isosceles triangle ABC where the vertex angle ∠A = 100°, we need to find the base angles. Using the angle sum property, the sum of all three angles in a triangle equals 180°. Since the two base angles are equal in an isosceles triangle, if ∠B = ∠C, then: 100° + ∠B + ∠C = 180°. This gives us 100° + 2∠B = 180°, so 2∠B = 80°, and therefore ∠B = ∠C = 40°.
In simple words: The two bottom angles of the triangle must be equal to each other. Together with the top angle of 100°, they add up to 180°, so each bottom angle is 40°.

Exam Tip: Remember that in any isosceles triangle, if you know the vertex angle, you can always find both base angles using the fact that they are equal and sum with the vertex angle to 180°.

Question 3. In ∆ABC, if AB = AC
Answer: Given that AB = AC, triangle ABC is isosceles with A as the vertex. The base angles ∠B and ∠C are equal. From the diagram, ∠B = 65°, so ∠C = 65°. Using the angle sum property: ∠A + ∠B + ∠C = 180°, we get ∠A + 65° + 65° = 180°, which gives ∠A = 180° - 130° = 50°.
In simple words: Since two sides are equal, the angles at the bottom are the same. Add them together with the top angle to get 180°.

Exam Tip: Mark the equal angles in the diagram immediately after identifying equal sides - this visual cue prevents careless errors.

Question 4. Let ABC be an isosceles triangle in which AB = AC. Then we have ∠B = ∠C. Let ∠B = ∠C = x. Then vertex angle A = 2(x + x) = 4x.
Answer: In isosceles triangle ABC with AB = AC, let the base angles be ∠B = ∠C = x. The vertex angle ∠A = 4x. By the angle sum property: ∠A + ∠B + ∠C = 180°, so 4x + x + x = 180°. This simplifies to 6x = 180°, giving x = 30°. Therefore, the vertex ∠A = 4 × 30° = 120°, and the base angles ∠B = ∠C = 30°.
In simple words: The top angle is four times bigger than each of the bottom angles. When you add all three angles together, they equal 180°.

Exam Tip: When angles are expressed as multiples of each other, always set up an equation using the angle sum property and solve systematically.

Question 5. In a right angled isosceles triangle, the vertex angle is ∠A = 90° and the other two base angles are equal. Let x° be the base angle and we have, ∠B = ∠C = 90°. By angle sum property of a triangle, we have
Answer: In a right angled isosceles triangle, the vertex angle ∠A = 90°. The two base angles ∠B and ∠C are equal. Let each base angle = x°. Using the angle sum property: 90° + x° + x° = 180°, which gives 90° + 2x° = 180°. Solving for x: 2x° = 90°, so x° = 45°. Therefore, ∠B = ∠C = 45°.
In simple words: The right angle is 90°, and the other two angles must be equal and add up to 90°, making each one 45°.

Exam Tip: A right angled isosceles triangle always has angles of 90° - 45° - 45°; memorizing this standard triangle helps solve problems faster.

Question 6. Given: ABC is an isosceles triangle in which AB = AC and BC is produced both ways.
Answer: In isosceles triangle ABC with AB = AC, the base angles are equal: ∠B = ∠C. When BC is extended beyond C to point D and beyond B to point E, exterior angles are formed. The exterior angle EBA equals the sum of the two non-adjacent interior angles: ∠EBA = ∠A + ∠C. Similarly, the exterior angle DCA equals ∠A + ∠B. Since ∠B = ∠C, we have exterior ∠EBA = exterior ∠DCA. This proves that the exterior angles on both sides are equal.
In simple words: When you extend the base of an isosceles triangle on both sides, the two outside angles you create are equal to each other.

Exam Tip: Always use the exterior angle theorem (exterior angle = sum of remote interior angles) in problems involving extended sides.

Question 7. Let be an equilateral triangle. Since it is an equilateral triangle, all the angles are equiangular and the measure of each angle is 60°. The exterior angle of ∠A is ∠BAF. The exterior angle of ∠B is ∠ABD. The exterior angle of ∠C is ∠ACE.
Answer: In an equilateral triangle, all three interior angles measure 60° each. At vertex A, the interior angle ∠A and the exterior angle ∠BAF form a linear pair, so they add up to 180°. Thus, ∠BAF = 180° - 60° = 120°. Similarly, at vertex B: ∠ABD = 180° - 60° = 120°, and at vertex C: ∠ACE = 180° - 60° = 120°. Each exterior angle of an equilateral triangle is 120°.
In simple words: Each inside angle of an equilateral triangle is 60°. When you extend any side to make an outside angle, that outside angle is always 120°.

Exam Tip: For any triangle, an exterior angle and its adjacent interior angle are supplementary (sum to 180°) - use this to find exterior angles quickly.

Question 8. Given: Two lines AB and CD intersect at O and O is the midpoint of AB and CD.
Answer: Since O is the midpoint of both AB and CD, we have AO = OB and CO = OD. In triangles AOC and BOD, these equal segments form corresponding sides. The angles ∠AOC and ∠BOD are vertically opposite angles, so they are equal. By the Side - Angle - Side (SAS) congruence criterion, triangle AOC is congruent to triangle BOD. From this congruence, the corresponding sides are equal, giving AC = BD. Additionally, since ∠CAO = ∠DBO (corresponding angles in congruent triangles), lines AC and BD are parallel when cut by transversal AB, making AC parallel to BD.
In simple words: When two line segments cross at their midpoints, the segments connecting the four endpoints are equal and parallel to each other.

Exam Tip: Identify the vertical angles immediately after marking the midpoints - they are your key to establishing congruence.

Question 9. Given: PA ⊥ AB, QB ⊥ AB, and PA = QB.
Answer: Since PA and QB are both perpendicular to AB, we have ∠PAO = ∠QBO = 90°. We are given that PA = QB. In triangles APO and BQO, these perpendiculars and equal segments give us PA = QB. The angle ∠PAO = ∠QBO = 90°. Since PA is perpendicular to AB and QB is perpendicular to AB, and both are transversals, the alternate angles formed are equal. By the Angle - Side - Angle (ASA) criterion, triangle APO is congruent to triangle BQO. From corresponding parts of congruent triangles, AO = OB and PO = OQ. This means O is the midpoint of both AB and PQ.
In simple words: When two perpendiculars of equal length are drawn from the ends of a line segment, the point where they meet is equidistant from both ends.

Exam Tip: Perpendicular lines create 90° angles - always highlight these right angles as they often unlock congruence proofs.

Question 10. Given: Line segments AB and CD intersect at O such that OA = OD and OB = OC.
Answer: To prove AC = BD, consider triangles AOC and BOD. We have OA = OD (given) and OC = OB (given). The angles ∠AOC and ∠BOD are vertically opposite, so ∠AOC = ∠BOD. By the Side - Angle - Side (SAS) congruence criterion, triangle AOC ≅ triangle BOD. From corresponding parts of congruent triangles, AC = BD. Additionally, since ∠CAO = ∠DBO (corresponding angles), lines AC and BD form equal alternate angles with transversal AB, which means AC is parallel to BD.
In simple words: When two intersecting line segments have matching pairs of equal parts, the line segments connecting opposite ends are equal and parallel.

Exam Tip: Always use vertically opposite angles in intersection problems - they are guaranteed equal and often the key to unlocking SAS congruence.

Question 11. Given: Two lines l and m are parallel to each other. M is the midpoint of segment AB. The line segment CD meets AB at M.
Answer: In triangles AMC and BMD, we need to show that M is the midpoint of CD (meaning CM = MD). Since l and m are parallel, and AB is a transversal, the alternate angles are equal: ∠MAC = ∠MBD. We are given AM = MB (M is the midpoint of AB). The angles ∠AMC and ∠BMD are vertically opposite angles, so ∠AMC = ∠BMD. By the Angle - Side - Angle (ASA) criterion, triangle AMC is congruent to triangle BMD. From corresponding parts of congruent triangles, CM = MD. Therefore, M is the midpoint of CD.
In simple words: If a line crosses through the midpoint of one segment between two parallel lines, it will also be cut in half at that same point.

Exam Tip: Use the parallel line property to get alternate angles immediately, then combine with the known midpoint to establish angle-side-angle congruence.

Question 12. Given: AB = AC and O is an interior point of the triangle such that OB = OC.
Answer: To prove ∠ABO = ∠ACO, consider triangles ABO and ACO. We are given AB = AC (the triangle is isosceles). We are also given OB = OC. The side AO is common to both triangles. By the Side - Side - Side (SSS) congruence criterion, triangle ABO is congruent to triangle ACO. From corresponding parts of congruent triangles, ∠ABO = ∠ACO.
In simple words: When a point inside an isosceles triangle is equidistant from the two equal sides' endpoints, it creates two triangles with equal angles at the base.

Exam Tip: The SSS criterion is powerful when all three pairs of sides are equal - check for common sides that might be overlooked at first glance.

Question 13. Given: A ∆ABC in which AB = AC and DE ∥ BC.
Answer: Since DE is parallel to BC and AB is a transversal, corresponding angles are equal: ∠ADE = ∠ABC (equation i). Similarly, since DE is parallel to BC and AC is a transversal, corresponding angles are equal: ∠AED = ∠ACB (equation ii). We are given AB = AC, which means the triangle is isosceles, so ∠ABC = ∠ACB (equation iii). From equations (i), (ii), and (iii), we have ∠ADE = ∠AED. In triangle ADE, since ∠ADE = ∠AED, the triangle is isosceles, which means AD = AE.
In simple words: A line parallel to the base of an isosceles triangle cuts the two equal sides into segments that are equal in length.

Exam Tip: Use the parallel line property to match angles, then apply the isosceles triangle property to show equal segments.

Question 14. Given: AX = AY.
Answer: In triangles AXC and AYB, we have AX = AY (given). The angle ∠A is common to both triangles. To apply the Side - Angle - Side (SAS) criterion, we need to show that AC = AB. Looking at the triangle ABC, if the configuration is symmetric or if additional conditions are given about the triangle, then we can establish AC = AB. With AX = AY, ∠A common, and AC = AB, by SAS criterion, triangle AXC is congruent to triangle AYB. From corresponding parts, XC = YB.
In simple words: When two segments from a vertex are equal, and the arms they lie on are also equal, the segments they cut off from those arms are equal to each other.

Exam Tip: Always identify the common angle and equal given sides first, then look for the third pair needed to complete the congruence criterion.

Question 15. Given: C is the mid point of a line segment AB, and D is point such that ∠DCA = ∠ECB and ∠DBC = ∠EAC.
Answer: In triangles ACE and DCB, we are given AC = BC (C is the midpoint of AB). We are also given ∠EAC = ∠DBC. From the given angle conditions and using properties of exterior angles, we can show ∠EAC = ∠DBC. Since the angle condition ∠DCA = ∠ECB relates the angles at C in both triangles, and applying the Angle - Side - Angle (ASA) criterion with the equal sides and corresponding angles, triangle ACE is congruent to triangle DCB. From corresponding parts of congruent triangles, DC = EC, which means DC = EC.
In simple words: When two triangles share conditions about equal angles and a common midpoint, their corresponding segments become equal.

Exam Tip: Midpoint conditions often provide equal segments - use these systematically with angle conditions to establish congruence.

Question 16. Given: AB ⊥ AC and DE ⊥ FE such that AB = DE and BF = CD.
Answer: In triangles ABC and DEF, we are given that ∠BAC = ∠DEF = 90° (both are right angles). We have AB = DE (given). From the relationships BC = BF + FC and FD = FC + CD, combined with the condition BF = CD (given), we can show BC = FD. By the Right angle - Hypotenuse - Side (RHS) congruence criterion, triangle ABC is congruent to triangle DEF. From corresponding parts of congruent triangles, AC = EF.
In simple words: Two right triangles with one equal leg and equal hypotenuses must have their other legs equal too.

Exam Tip: The RHS criterion applies only to right triangles - always verify the 90° angle before using this powerful congruence rule.

Question 17. Given: AB = BC and x° = y°.
Answer: In triangles BCD and BAE, the exterior angle of triangle ABE at B is y° = ∠EBA + ∠BAE. In triangle BCD, the exterior angle is x° = ∠CBA + ∠BCD. Since x° = y° (given), we have ∠EBA + ∠BAE = ∠CBA + ∠BCD. We are given AB = BC. The common angle ∠B is the same in both triangles. With AB = BC, the common angle at B, and the equal exterior angle condition, by the Angle - Side - Angle (ASA) criterion, triangle BCD is congruent to triangle BAE. From corresponding parts of congruent triangles, CD = AE.
In simple words: When two segments from a point are equal and create equal exterior angles, the segments they determine on extended lines are equal.

Exam Tip: Exterior angle equality is a subtle condition - always break it down into its component interior angles using the exterior angle theorem.

Question 18. Given: A triangle ABC in which AB = AC and BD and CE are the bisectors of angle B and angle C respectively. To prove: BD = CE
Answer: In triangles ABD and ACE, we have angle ABD equals half of angle B (since BD bisects angle B), and angle ACE equals half of angle C (since CE bisects angle C). Because AB = AC, angle B equals angle C (base angles of an isosceles triangle are equal). Therefore, angle ABD equals angle ACE. We also know AB = AC (given) and angle BAD is common to both triangles. By the Angle-Side-Angle congruence criterion, triangle ABD is congruent to triangle ACE. Since the corresponding parts of congruent triangles are equal, we have BD = CE.
In simple words: When two sides of a triangle are equal, and we draw angle bisectors from the base angles, those bisectors turn out to be the same length. This happens because the triangles formed are congruent.

Exam Tip: Always identify which triangles you are comparing and list all three conditions for congruence (ASA, SAS, SSS) carefully before concluding the proof.

Question 19. Given: A triangle ABC in which D is the mid point of BC and BL is perpendicular to AD and CM is perpendicular to AD. To prove: BL = CM
Answer: In triangles BLD and CMD, we have angle BLD equals angle CMD equals 90 degrees (given). Angle BDL equals angle MDC (vertically opposite angles). And BD equals DC (since D is the midpoint of BC, given). By the Angle-Angle-Side congruence criterion, triangle BLD is congruent to triangle CMD. The corresponding parts of congruent triangles are equal, so BL equals CM.
In simple words: When you drop perpendiculars from two equal segments to the same line, the lengths of those perpendiculars will be the same.

Exam Tip: Mark the right angles and equal segments clearly on the diagram - this makes spotting the congruent triangles much easier.

Question 20. Given: In a triangle ABC, D is the mid point of BC and DL is perpendicular to AB and DM is perpendicular to AC. Also, DL = DM. To prove: AB = AC
Answer: In right-angled triangles BLD and CMD, we have angle BLD equals angle CMD equals 90 degrees. The hypotenuse BD equals the hypotenuse CD (given, since D is the midpoint). And DL equals DM (given). By the Right Angle-Hypotenuse-Side criterion of congruence, triangle BLD is congruent to triangle CMD. The corresponding parts of congruent triangles are equal, so angle ABD equals angle ACD. In triangle ABC, since angle ABD equals angle ACD, the sides opposite to these equal angles are equal. Therefore, AB equals AC (sides opposite to equal angles are equal).
In simple words: If you drop equal perpendiculars from a point to two sides of a triangle, and that point is equidistant from the endpoints of the base, then the two sides must be equal in length.

Exam Tip: Use the RHS criterion when you have a right angle, equal hypotenuses, and one other equal side - this is a powerful shortcut.

Question 21. Given: A triangle ABC in which AB = AC, BO and CO are bisectors of angle B and angle C
Answer: In triangle BOC, we can show angle OBC equals half of angle B and angle OCB equals half of angle C. Since AB equals AC, angle B equals angle C (base angles of an isosceles triangle). Therefore, angle OBC equals angle OCB, which means OB equals OC (sides opposite to equal angles are equal). In triangles ABO and ACO, we have AB equals AC (given), angle ABO equals angle ACO (from the above), and BO equals OC (proved above). By the Side-Angle-Side criterion of congruence, triangle ABO is congruent to triangle ACO. The corresponding parts of congruent triangles are equal, so angle BAO equals angle CAO. This shows that AO bisects angle A.
In simple words: In an isosceles triangle, if you draw the angle bisectors from the two equal base angles, they create smaller triangles that are congruent to each other, which leads to the bisector from the top angle.

Exam Tip: Recognize that angle bisectors in isosceles triangles often lead to equal angles and equal sides - use this pattern to spot congruent triangles quickly.

Question 22. Given: PQR is an equilateral triangle and QRST is a square. To prove: PT = PS and angle PSR = 15°
Answer: Since triangle PQR is equilateral, angle PQR equals 60 degrees and angle PRQ equals 60 degrees. Since QRST is a square, angle QRT equals 90 degrees and angle QRS equals 90 degrees. In triangle PQT, angle PQT equals angle PQR plus angle QRT, which is 60 degrees plus 90 degrees, equalling 150 degrees. In triangle PRS, angle PRS equals angle PRQ plus angle QRS, which equals 60 degrees plus 90 degrees, equalling 150 degrees. Therefore, angle PQT equals angle PRS. Since PQ equals PR (sides of equilateral triangle PQR), and QT equals RS (sides of square QRST), by the Side-Angle-Side criterion of congruence, triangle PQT is congruent to triangle PRS. The corresponding parts of congruent triangles are equal, so PT equals PS. In triangle PRS, since PR equals RS and angle PRS equals 150 degrees, we use the angle sum property. Since PR equals RS, triangle PRS is isosceles, so angle RPS equals angle RSP. We have angle PRS plus angle RPS plus angle RSP equals 180 degrees, giving 150 degrees plus 2 times angle RSP equals 180 degrees. Solving, angle RSP equals 15 degrees.
In simple words: When you attach a square to one side of an equilateral triangle, the distances from the far corners of the square to the opposite vertex of the triangle are equal. The angle at one corner of the square facing the triangle works out to exactly 15 degrees.

Exam Tip: Break complex figures into simpler components - identify the equilateral triangle and square separately, then use angle properties to find unknown angles.

Question 23. Given: ABC is a triangle, right angled at B. ACFG is a square and BCDE is a square. To prove: AD = EF
Answer: Since BCDE is a square, angle BCD equals 90 degrees. In angle ACD, we have angle ACD equals angle ACB plus angle BCD, which equals angle ACB plus 90 degrees. In triangle BCF, angle BCF equals angle BCA plus angle ACF. Since ACFG is a square, angle ACF equals 90 degrees. So angle BCF equals angle BCA plus 90 degrees. From equations (2) and (3), we see angle ACD equals angle BCF. Thus, in triangles ACD and BCF, we have AC equals CF (sides of square ACFG), angle ACD equals angle BCF (proved above), and CD equals BC (sides of square BCDE). By the Side-Angle-Side criterion of congruence, triangle ACD is congruent to triangle BCF. The corresponding parts of congruent triangles are equal, so AD equals BF.
In simple words: When you build squares on two perpendicular sides of a right triangle and on the hypotenuse, certain distances from corners of these squares back to vertices of the triangle turn out to be equal.

Exam Tip: In problems involving squares attached to triangles, pay close attention to angle sums - they often produce equal angles in different parts of the figure.

Question 24. Given: ABC is an isosceles triangle in which AB = AC and AD is the median through A. To prove: angle BAD = angle CAD
Answer: In triangles ABD and ACD, we have AB equals AC (given). BD equals DC (given, since AD is the median, D is the midpoint of BC). AD equals AD (common side). By the Side-Side-Side criterion of congruence, triangle ABD is congruent to triangle ACD. The corresponding parts of congruent triangles are equal, so angle BAD equals angle CAD.
In simple words: In an isosceles triangle, the line from the top vertex to the midpoint of the base divides the top angle into two equal parts.

Exam Tip: Remember that in an isosceles triangle, the median from the vertex angle to the base is also the angle bisector and the altitude - these coincide.

Question 25. Given ABCD is a quadrilateral in which AB is parallel to DC. To prove: (i) AB = CQ (ii) DQ = DC + AB
Answer: (i) In triangles ABP and PCQ, we have angle PAB equals angle PQC (alternate angles, since AB is parallel to DC). Angle APB equals angle CPQ (vertically opposite angles). And BP equals PC (given). By the Angle-Angle-Side criterion of congruence, triangle ABP is congruent to triangle PCQ. The corresponding parts of congruent triangles are equal, so AB equals CQ. (ii) Now, DQ equals DC plus CQ. From part (i), CQ equals AB. Therefore, DQ equals DC plus AB.
In simple words: When two sides of a quadrilateral are parallel and you join certain vertices, the resulting segments follow a simple additive relationship.

Exam Tip: In parallel line problems, look for alternate angles - they are usually equal and help prove triangle congruence.

Question 26. Given: OA = OB and OP = OQ
Answer: (i) In triangles OAQ and OPB, we have OA equals OB (given). Angle O equals angle O (common). And OQ equals OP (given). By the Side-Angle-Side criterion of congruence, triangle OAQ is congruent to triangle OPB. The corresponding parts of congruent triangles are equal, so angle OBP equals angle OAQ. In triangles BXQ and PXA, we have angle BXQ equals angle PXA (vertically opposite angles). Angle OBP equals angle OAQ (from above). And BQ equals PA (can be shown from the congruence). By the Angle-Angle-Side criterion of congruence, triangle BXQ is congruent to triangle PXA. The corresponding parts of congruent triangles are equal, so PX equals QX and AX equals BX.
In simple words: When two rays from a point have equal distances marked on them, and perpendiculars are drawn to each ray, the intersection point of these perpendiculars creates equal segments.

Exam Tip: Use SAS carefully - make sure the angle is between the two sides you are comparing for equality.

Question 27. Given: ABCD is a square and P is a point inside it such that PB = PD
Answer: In triangles APD and APB, we have DA equals AB (since ABCD is a square). AP equals AP (common side). And PB equals PD (given). By the Side-Side-Side criterion of congruence, triangle APD is congruent to triangle APB. The corresponding parts of congruent triangles are equal, so angle DAP equals angle BAP. Similarly, in triangles CPD and CPB, we have CD equals CB (sides of square). CP equals CP (common). And PB equals PD (given). By the Side-Side-Side criterion, triangle CPD is congruent to triangle CPB. Therefore, angle DCP equals angle BCP. Adding the angles, angle DAP plus angle DCP equals angle BAP plus angle BCP. From the congruence results, angle DAP equals angle BAP, and since angle APD equals angle APB (from earlier congruence), we can show that C, P, and A are collinear, proving CPA is a straight line.
In simple words: If a point inside a square is equidistant from two opposite corners, then the line joining the other two opposite corners passes through this point.

Exam Tip: Use the property that if two triangles are congruent, their corresponding angles are equal - this helps build the case for collinearity.

Question 28. A triangle ABC which is an equilateral triangle and PQ is parallel to AC. AC is produced to R such that CR = BP
Answer: Let QR intersect PC at M. Since triangle ABC is equilateral, angle A equals angle ACB equals 60 degrees. Since PQ is parallel to AC and corresponding angles are equal, angle BPQ equals angle BAC, which is 60 degrees. In triangle BPQ, angle B equals angle ACB equals 60 degrees (since triangle ABC is equilateral), and angle BPQ equals 60 degrees. Therefore, angle BQP equals 60 degrees, making triangle BPQ equilateral. This gives PQ equals BP equals BQ. Since BP equals CR (given), we have PQ equals CR. Consider triangles PMQ and CMR. Since PQ is parallel to AC, angle PMQ equals angle CMR (vertically opposite). Angle PQM equals angle CRM (alternate angles). And PQ equals CR (proved above). By the Angle-Angle-Side criterion of congruence, triangle PMQ is congruent to triangle CMR. The corresponding parts of congruent triangles are equal, so PM equals CM.
In simple words: In an equilateral triangle, when you draw a parallel line to one side and extend the triangle in a specific way, the intersection points create equal segments.

Exam Tip: Equilateral triangles have all angles equal to 60 degrees - use this repeatedly to identify other 60-degree angles in the figure.

Question 29. Given: a quadrilateral ABCD in which AB = AD and BC = DC
Answer: In triangles ABC and ADC, we have AB equals AD (given). BC equals DC (given). And AC equals AC (common side). By the Side-Side-Side criterion of congruence, triangle ABC is congruent to triangle ADC. The corresponding parts of congruent triangles are equal, so angle BAC equals angle DAC, which means AC bisects angle A. Also, angle BCA equals angle DCA, which means AC bisects angle C. For part (ii), in triangles ABO and ADO, we have AB equals AD (given). Angle BAO equals angle DAO (since AC bisects angle A). And AO equals AO (common). By the Side-Angle-Side criterion of congruence, triangle ABO is congruent to triangle ADO. The corresponding parts of congruent triangles are equal, so angle BAO equals angle DAO and BO equals DO. Since angle BOA plus angle DOA equals 180 degrees (as shown in the congruence), we get angle BOA equals 90 degrees. This means AC is perpendicular to BD, and furthermore, AC bisects BD.
In simple words: In a quadrilateral where two pairs of adjacent sides are equal, the diagonal connecting the vertices between these pairs acts as a perpendicular bisector of the other diagonal.

Exam Tip: Kite-shaped quadrilaterals (with two pairs of equal adjacent sides) always have perpendicular diagonals - one diagonal bisects the other.

Question 30. Given: A triangle ABC in which bisectors of angle B and angle C meet at I. Also, IP is perpendicular to BC, IQ is perpendicular to CA and IR is perpendicular to AB
Answer: (i) In triangles BIP and BIR, we have angle PBI equals angle RBI (since BI bisects angle B). Angle BPI equals angle BRI equals 90 degrees (given). And BI equals BI (common). By the Angle-Angle-Side criterion of congruence, triangle BIP is congruent to triangle BIR. The corresponding parts of congruent triangles are equal, so IP equals IR. Similarly, by considering triangles CIQ and CIP, and using the fact that CI bisects angle C, we can show IP equals IQ. Therefore, IP equals IQ equals IR. (ii) In triangles AIR and AIQ, we have IR equals IQ (proved above). IA equals IA (common). And angle IRA equals angle IQA equals 90 degrees (given). By the Side-Angle-Side criterion of congruence, triangle AIR is congruent to triangle AIQ. The corresponding parts of congruent triangles are equal, so angle IAR equals angle IAQ. This means IA bisects angle A.
In simple words: The point where any two angle bisectors of a triangle meet is equidistant from all three sides of the triangle, and the line from this point to the third vertex also bisects that angle.

Exam Tip: The incenter (where all three angle bisectors meet) is the center of the inscribed circle - remember that all perpendicular distances from the incenter to the sides are equal.

Question 31. Given: An angle AOB and P is a point in the interior of angle AOB such that PL = PM. Also PL = OA and PM = OB
Answer: In triangles OPL and OPM, we have angle OMP equals angle OLP equals 90 degrees (given). OP equals OP (common side). And PL equals PM (given). By the Right Angle-Hypotenuse-Side criterion of congruence, triangle OPL is congruent to triangle OPM. The corresponding parts of congruent triangles are equal, so angle POL equals angle POM. This means OP is the bisector of angle LOM, which equals angle AOB.
In simple words: If a point inside an angle is such that its perpendicular distances to the two rays are equal, then the line from the vertex to this point bisects the angle.

Exam Tip: Equal perpendicular distances from a point to two lines always indicate that the point lies on the angle bisector - this is a fundamental property of angle bisectors.

Question 32. Given M is the mid-point of side AB of a square ABCD and CM ⊥ PQ
Answer: In triangles AMP and BMQ, angles AMP and BMQ are vertically opposite, so they are equal. Since ABCD is a square, angles PAM and MBQ both measure 90°. Also, AM equals MB (given that M is the midpoint). By the Angle-Angle-Side criterion, triangle AMP is congruent to triangle BMQ. From this congruence, PA equals BQ and MP equals MQ.

In triangles PCM and QCM, we have PM equals QM (from the congruence above). Since CM ⊥ PQ (given), angles PMC and QMC are both 90°. CM is common to both triangles. By the Side-Angle-Side criterion, triangle PCM is congruent to triangle QCM. From this congruence, PC equals QC, which gives us PC = QB + CB. Since AB equals CB and PA equals QB, we get PC = AB + PA.
In simple words: Two pairs of triangles become congruent - first AMP with BMQ, then PCM with QCM. Using these congruences in steps, we show that PC ends up equaling AB plus PA.

Exam Tip: Mark all equal segments and angles from the given conditions (midpoint, perpendicularity, square properties) before starting the proof - this clarifies which congruence criterion to apply.

Question 33. Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point A, O and N lie on a straight line where point O is the mid point of BM.
Answer: In triangles ABO and NMO, angles ABO and OMN both equal 90°. Since O is the midpoint of BM, we have OB = OM. The angles AOB and MON are vertically opposite angles. By the Angle-Side-Angle criterion, triangle ABO is congruent to triangle NMO. From this congruence of the two triangles, their matching parts are equal, giving us AB = NM. This shows that MN represents the breadth of the river.
In simple words: By setting up the perpendicular and using the fact that O is the midpoint, two triangles become congruent. This means AB and MN must have the same length, so measuring MN on land tells us the river's width.

Exam Tip: Always identify vertically opposite angles in geometry problems - they are automatically equal without needing any additional justification.

Question 34. We have ∠A = 36° and ∠B = 64°. By the angle sum property in △ABC, we have ∠A + ∠B + ∠C = 180°. ⇒ 36° + 64° + ∠C = 180°. ⇒ ∠C = 180° - 100° = 80°. Therefore, we have ∠A = 36°, ∠B = 64° and ∠C = 80°. ∠C is largest and ∠A is shortest. Side opposite to ∠C is longest and hence AB is longest side. Side opposite to ∠A is shortest and hence BC is shortest side.
Answer: The three angles of the triangle total 180°. When we add the two given angles (36° + 64°), we get 100°. The remaining angle must therefore be 80°. Comparing all three angles, the 80° angle is the largest while 36° is the smallest. In any triangle, the longest side is always opposite the largest angle, and the shortest side lies opposite the smallest angle. Therefore, side AB (opposite the 80° angle) is the longest, and side BC (opposite the 36° angle) is the shortest.
In simple words: Use the angle sum rule to find the third angle. Then remember: bigger angle - longer opposite side, smaller angle - shorter opposite side.

Exam Tip: Always list all three angles in order before identifying longest and shortest sides - this prevents careless mistakes in angle comparison.

Question 35. In a right angle triangle, greatest angle is ∠A = 90°. And hence other angles are less than 90° because sum of the angles of a triangle is 180°. So, ∠A is the greatest angle. Therefore, side BC which is opposite to ∠A is longest.
Answer: A right-angled triangle has one 90° angle, which is the largest possible angle in that triangle. All other angles must be less than 90° (since the three angles sum to 180°). The side that lies opposite to the largest angle is always the longest side. Therefore, the side opposite the 90° angle, which is BC, must be the longest.
In simple words: The right angle (90°) is always the biggest angle in a right triangle. The side across from it is therefore always the longest side.

Exam Tip: In any right triangle, the side opposite the right angle (called the hypotenuse) is always the longest - remember this as a standard fact.

Question 36. In △ABC, ∠A = ∠B = 45°. So, ∠C = 180° - ∠A - ∠B = 180° - 45° - 45° = 180° - 90° = 90°. Thus we find that ∠C is the greatest angle of △ABC. So, AB is the longest side which is opposite to ∠C.
Answer: Given that two angles are each 45°, the third angle works out to 90° when we apply the angle sum property. The 90° angle is therefore the largest angle in the triangle. Since the longest side of any triangle always faces the largest angle, side AB (which is opposite the 90° angle at C) must be the longest side of triangle ABC.
In simple words: Two 45° angles leave 90° for the third angle. That 90° angle is the biggest, so the side opposite it is the longest.

Exam Tip: For isosceles right triangles (two 45° angles), quickly recall that the third angle is always 90° and the side opposite it is the hypotenuse (longest side).

Question 37. In △ABC, ∠A + ∠B + ∠C = 180°. ⇒ 70° + 60° + ∠C = 180°. ⇒ 130° + ∠C = 180°. ⇒ ∠C = 180° - 130° = 50°. Now in △BCD we have, ∠CBD = ∠DAC + ∠ACB [∵ ∠CBD is the exterior angle of ∠ABC] = 70° + 50° = 120°. Since BC = BD [Given]. So, ∠BCD = ∠BDC. ∴ ∠BCD + ∠BDC = 180° - ∠CBD = 180° - 120° = 60°. ⇒ 2∠BCD = 60°. ⇒ ∠BCD = ∠BDC = 30°. Now in △ACD we have. ∠A = 70°, ∠D = 30° and ∠ACD = ∠ACB + ∠BCD = 50° + 30° = 80°. ∴ ∠ACD is the greatest angle. So the side opposite to ∠ACD, that is AD, is the longest side of △ACD. ∴ AD > CD. (ii) Since ∠BDC is the smallest angle, the side opposite to ∠BDC, that is AC, is the shortest side of △ACD. ∴ AD > AC.
Answer: First, find angle C in triangle ABC using the angle sum property: 180° - 70° - 60° = 50°. Since BC = BD (given), the base angles in triangle BCD are equal. The exterior angle CBD equals 70° + 50° = 120°. This gives us equal angles BCD and BDC, each measuring 30° (since they sum to 60°). In triangle ACD, the angles are now 70°, 30°, and 80°. The 80° angle at C is the largest, so the opposite side AD is the longest in triangle ACD. Therefore AD exceeds CD. Similarly, the 30° angle at D is the smallest in triangle ACD, making AC (opposite to it) the shortest side. This means AD exceeds AC.
In simple words: Use the angle sum rule and the isosceles property (equal sides give equal base angles) to find all angles in the smaller triangle. Then apply the rule: bigger angle gets the longer opposite side.

Exam Tip: When an isosceles condition is given (like BC = BD), immediately mark the two equal base angles - this unlocks the rest of the angle calculations quickly.

Question 38. In △ABC, ∠A = 180° - ∠B - ∠C = 180° - 35° - 65° = 180° - 100° = 80°. ∴ ∠BAX = (1/2)∠A = (1/2) × 80° = 40°. Now in △ABX, ∠B = 35°, ∠BAX = 40° and ∠BXA = 180° - 35° - 40° = 180° - 75° = 105°. So, in △ABX, ∠B is smallest, so the side opposite to ∠B, that is AX, is smallest. So AX < BX ....(i). Now consider △AXC. ∠CAX = (1/2) × ∠A = (1/2) × 80° = 40°. ∠AXC = 180° - 40° - 65° = 180° - 105° = 75°. Therefore, in △AXC, we have, ∠CAX = 40°, ∠C = 65° and ∠AXC = 75°. ∴ ∠CAX is smallest in △AXC. So the side opposite to ∠CAX is shortest. ⇒ CX is shortest. ⇒ CX < AX ....(ii). From (i) and (ii), we get BX > AX > CX. This is the required descending order.
Answer: The third angle of triangle ABC works out to 80° using the angle sum property. Since the line AX bisects angle A, each part equals 40°. In triangle ABX, angle B (35°) is the smallest, making AX (opposite to it) the smallest side in that triangle. So AX is less than BX. In triangle AXC, angle CAX (40°) is the smallest, making CX the shortest side in that triangle. So CX is less than AX. Combining these two results gives the order: BX is greater than AX, which is greater than CX.
In simple words: An angle bisector splits angle A into two 40° pieces. Then use the "smallest angle has the shortest opposite side" rule twice - once for each of the two smaller triangles formed.

Exam Tip: When an angle bisector is drawn, calculate the two half-angles first, then analyze each resulting triangle separately before combining results.

Question 39. Given: ABC is a triangle in which AD is the bisector of ∠A. Proof: (i) In △ACD, the exterior angle ADB equals ∠DAC + ∠ACD, which simplifies to ∠BAD + ∠ACD (since ∠DAC = ∠BAD, given). Therefore, ∠ADB is greater than ∠BAD. The side opposite to angle ADB is the longest side in △ADB. So, AB is longer than BD. (ii) Again in △ABD, the exterior angle ADC equals ∠ABD + ∠BAD, which gives ∠ABD + ∠CAD. Therefore, ∠ADC is greater than ∠CAD. The side opposite to angle ADC is the longest side in △ACD. So, AC is longer than DC.
Answer: When the angle bisector AD is drawn, it splits angle A into two equal parts. Looking at triangle ABD, the exterior angle ADB (formed by extending side AD) equals the sum of the two non-adjacent interior angles. This exterior angle is larger than angle BAD (half of angle A). Since the larger angle has the longer opposite side, AB must be longer than BD. Similarly, in triangle ACD, the exterior angle ADC is larger than angle CAD. This means AC is longer than DC.
In simple words: An angle bisector creates an exterior angle in each of the two smaller triangles. Each exterior angle is larger than the adjacent interior angle, which tells us about which sides are longer.

Exam Tip: Always use the exterior angle theorem (exterior angle equals sum of remote interior angles) - it directly compares angles and helps identify the longest sides.

Question 40. Given: △ABC is which AB = AC side BC of △ABC is produced to D. To prove: AD > AC. Proof: In △ABC, the exterior angle ACD (formed when BC is extended to D) equals ∠B + ∠BAC. Since AB equals AC (given), angles ∠C and ∠B are equal. Therefore, the exterior angle ACD equals ∠ACB + ∠BAC. The exterior angle ACD is greater than angle CDA. So the side opposite to angle ACD is the longest. Therefore, AD is greater than AC.
Answer: The triangle ABC has two equal sides AB and AC. When BC is extended to point D, an exterior angle is formed at C. By the exterior angle property, this angle equals the sum of angles B and BAC. Since AB = AC, the triangle is isosceles, meaning angles B and C are equal. The exterior angle at C is therefore larger than any interior angle of the triangle. The side opposite the exterior angle (which is AD) must be longer than the side opposite angle CAD or angle ADC. Thus AD exceeds AC.
In simple words: When you extend one side of an isosceles triangle, the new exterior angle is bigger than the original angles. The opposite side to this big angle must be the longest.

Exam Tip: For isosceles triangles, remember that equal sides have equal opposite angles - use this symmetry to compare exterior angles with interior angles.

Question 41. Given: △ABC in which AC > AB and AD is a bisector of ∠A. To prove: ∠ADC > ∠ADB. Proof: Since AC > AB, we have ∠ABC > ∠ACB. Adding (1/2)∠A on both sides of this inequality, ∠ABC + (1/2)∠A > ∠ACB + (1/2)∠A. ⇒ ∠ABC + ∠Z BAD > ∠ACB + ∠DAC (∵ AD is a bisector of ∠A). ⇒ Exterior ∠ADC > Exterior ∠ADB. ∴ ∠ADC > ∠ADB.
Answer: Since side AC is longer than side AB, the angle opposite the longer side must be larger. This means angle ABC (opposite AC) is greater than angle ACB (opposite AB). Now add half of angle A to both sides of this inequality. This gives: angle ABC plus half-angle A is greater than angle ACB plus half-angle A. Since AD bisects angle A, each half equals ∠BAD and ∠DAC. The left side of the inequality becomes the exterior angle ADC, and the right side becomes the exterior angle ADB. Therefore, exterior angle ADC is greater than exterior angle ADB.
In simple words: Longer side means bigger opposite angle. Add equal amounts to both sides of the angle inequality, and you get a new inequality comparing two exterior angles.

Exam Tip: When dealing with inequalities in geometry, maintain the direction of the inequality while adding equal quantities to both sides - this preserves the comparison.

Question 42. Given: A triangle PQR and S is a point on QR. To prove: PQ + QR + RP > 2PS. Proof: Since in a triangle, sum of any two sides is always greater than the third side. So in △PQS, we have PQ + QS > PS ....(i). Similarly, in △PSR, we have PR + SR > PS ....(ii). Adding both sides of (i) and (ii), we get PQ + QS + PR + SR > 2PS. ⇒ PQ + PR + QS + SR > 2PS. ⇒ PQ + PR + QR > 2PS (∵ QS + SR = QR)
Answer: The triangle inequality states that for any triangle, the sum of two sides is always greater than the third side. Applying this to triangle PQS gives: PQ + QS exceeds PS. Applying the same rule to triangle PSR gives: PR + SR exceeds PS. When we add these two inequalities together, we get: PQ + QS + PR + SR exceeds 2PS. Since QS and SR together make up the full side QR, we can write: PQ + PR + QR exceeds 2PS.
In simple words: Break the original triangle into two smaller triangles by drawing a line from P to a point on the opposite side. Apply the triangle inequality to each piece, then combine the results.

Exam Tip: Always use the triangle inequality for any three points - it works for parts of triangles too, not just complete triangles.

Question 43. Given: A circle with centre O is drawn in which XY is a diameter and XZ is a chord. To prove: XY > XZ. Proof: In △XOZ, we have, OX + OZ > XZ [∵ sum of any two sides in a triangle is a greater than its third side]. ⇒ OX + OY > XZ [∵ OZ = OY, radius of the circle]. ∴ XY > XZ [∵ OX + OY = XY]
Answer: Consider triangle XOZ, where O is the centre. By the triangle inequality, the sum of two sides OX and OZ is greater than the third side XZ. Since both OX and OZ are radii, they equal the radius length. Therefore, OX + OY (where Y is another point on the circle) equals the diameter XY. This gives us: the diameter XY is greater than the chord XZ.
In simple words: A diameter connects two points on a circle through the centre. Any chord connects two points on the circle but does not pass through the centre. The triangle inequality guarantees that the longer path (through the centre) is always the diameter.

Exam Tip: For circle problems, remember that all radii are equal - this symmetry often helps simplify proofs using the triangle inequality.

Question 44. Given: ABC is a triangle and O is a point inside it. To Prove: (i) AB + AC > OB + OC, (ii) AB + BC + CA > OA + OB + OC, (iii) OA + OB + OC > ½(AB + BC + CA). Proof: (i) In △ABC, AB + AC > BC ....(i). And in △OBC, OB + OC > BC ....(ii). Subtracting (ii) from (i) we get (AB + AC) - (OB + OC) > (BC - BC), i.e. AB + AC > OB + OC. (ii) AB + AC > OB + OC [proved in (i)]. Similarly, AB + BC > OA + OC. And AC + BC > OA + OB. Adding both sides of these three inequalities, we get (AB + AC) + (AC + BC) + (AB + BC) > OB + OC + OA + OB + OA + OC, i.e. 2(AB + BC + AC) > 2(OA + OB + OC). Therefore, we have AB + BC + AC > OA + OB + OC. (iii) In △OAB, OA + OB > AB ....(i). In △OBC, OB + OC > BC ....(ii). And, in △OCA, OC + OA > CA. Adding (i), (ii) and (iii) we get (OA + OB) + (OB + OC) + (OC + OA) > AB + BC + CA, i.e 2(OA + OB + OC) > AB + BC + CA. ⇒ OA + OB + OC > ½(AB + BC + CA)
Answer: For part (i): In triangle ABC, the sum of two sides AB and AC exceeds BC by the triangle inequality. In triangle OBC, the sum OB + OC also exceeds BC. Subtracting the second inequality from the first removes BC and leaves AB + AC greater than OB + OC. For part (ii): Using the result from part (i) and applying similar reasoning to the other pairs of sides (AB with O, and AC with O), we get three inequalities. Adding them all gives 2(AB + BC + AC) greater than 2(OA + OB + OC), which simplifies to the required result. For part (iii): Apply the triangle inequality to each of the three triangles OAB, OBC, and OCA. Each gives a sum of two segments from O that exceeds the opposite side of the main triangle. Adding all three gives 2(OA + OB + OC) greater than the full perimeter, which rearranges to the final inequality.
In simple words: A point inside a triangle divides it into three smaller triangles. Use the triangle inequality on the main triangle and smaller triangles to build chains of inequalities that relate the perimeter to the distances from the interior point.

Exam Tip: When multiple inequalities must be added together, first align them so that cancellations are clear - this prevents sign errors.

Question 45. Since AB = 3 cm and BC = 3.5 cm, ∴ AB + BC = (3 + 3.5) cm = 6.5 cm. And CA = 6.5 cm. So AB + BC = CA. A triangle can be drawn only when the sum of two sides is greater than the third side. So, with the given lengths a triangle cannot be drawn.
Answer: The sum of two sides is 3 + 3.5 = 6.5 cm, which exactly equals the third side. For a valid triangle to exist, the sum of any two sides must be strictly greater than the third side. Since the sum equals (rather than exceeds) the third side, the three points would be collinear (lie on a straight line) rather than form a triangle. Therefore, no triangle can be constructed with these measurements.
In simple words: When two sides add up to exactly the length of the third side, the three points would form a flat line, not a triangle. You always need the sum to be bigger than the third side.

Exam Tip: The triangle inequality requires strictly greater than (>) not greater than or equal to (≥) - remember that equality means the points are collinear.