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Finish Line & Beyond CONGRUENCE OF TRIANGLES 1. Two figures are congruent, if they are of the same shape and of the same size. 2. Two circles of the same radii are congruent. 3. Two squares of the same sides are congruent. 4. If two triangles ABC and PQR are congruent under the correspondence A – P, B-Q and C-R, then symbolically, it is expressed as Δ ABC Δ PQR. 5. SAS Congruence Rule: If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent. (Axiom: This result cannot be proved with the help of previously known results.) 6. ASA Congruence Rule: If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent (ASA Congruence Rule). Construction: Two triangles are given as follows, where ABC DEF ACB DFE . Sides AB=DE To Prove: ABC DEF and Proof: ABC DEF (given) AB = DE AC= DF (Sides opposite to corresponding angles are in the same ratio as ratio of angles) Hence, by SAS congruence rule ABC DEF is proved. A B C www.excellup.com ©2009 send your queries to [email protected] Finish Line & Beyond D E F 7. AAS Congruence Rule: If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent. This theorem can be proved in similar way as the previous one. 8. Angles opposite to equal sides of a triangle are equal. 9. Sides opposite to equal angles of a triangle are equal. 10. Each angle of an equilateral triangle is of 60°. 11. SSS Congruence Rule: If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent. 12. RHS Congruence Rule: If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS Congruence Rule). 13. In a triangle, angle opposite to the longer side is larger (greater). 14. In a triangle, side opposite to the larger (greater) angle is longer. 15. Sum of any two sides of a triangle is greater than the third side. Theorem: Angles opposite to equal sides of an isosceles triangle are equal. Theorem: The sides opposite to equal angles of a triangle are equal. Theorem: If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater). Theorem: In any triangle, the side opposite to the larger (greater) angle is longer. Theorem: The sum of any two sides of a triangle is greater than the third side. www.excellup.com ©2009 send your queries to [email protected] Finish Line & Beyond EXERCISE 1 C 1. In quadrilateral ACBD, AC = AD and AB bisects A. Show that Δ ABC Δ ABD. Answer: In ACB & ADB AC=AD B A CAB DAB (AB is bisecting CAD ) AB = AB (common side in both triangles) D So, by SAS axiom it is proved that; ACB ADB 2. ABCD is a quadrilateral in which AD = BC and DAB = CBA . Prove that (i) Δ ABD Δ BAC (ii) BD = AC (iii) ABD = BAC. D A Answer: (i) In ABD & BAC AD=BC AB=AB (common) B C BAD ABC So, by SAS rule ABD BAC (ii) Since, ABD BAC , so BD = AC (third corresponding sides of respective triangles). B C (iii) In congruent triangles all corresponding angles Are always equal, so BAD ABC is proved 3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB. O D Answer: In BOC & AOD BC=AD (given) CBO DAO (Right Angle) BOC AOD (Opposite angles of intersecting lines So, by ASA rule BOC AOD BO AO and it is proved that CD bisects AB. www.excellup.com ©2009 send your queries to [email protected] A Finish Line & Beyond 4. l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that Δ ABC Δ CDA. p q A Answer: In ABC & CDA AB = CD (l and m are parallel) AD=BC (AB and CD are parallel) D B ABC DCm l m C (Angles on the same side of transversal BC) DCm ADC (Alternate Angles are equal) So, ABC ADC So, by SAS rule ABC CDA 5. Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A . Show that: (i) Δ APB Δ AQB (ii) BP = BQ or B is equidistant from the arms of A . Q l B A P Answer: In APB & AQB AB=AB (Common side) PAB QAB (AB is bisector of QAP ) AQB APB (Right Angle) So, by ASA rule APB AQB E And BQ=BP A 6. In the given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. Answer: In ABC & ADE AB=AD (given) AC=AE (given) Since, BAD EAC B D C So, BAD DAC EAC DAC Or, BAC DAE So, by SAS rule ABC ADE D E BC DE proved 7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB A www.excellup.com ©2009 send your queries to [email protected] P B Finish Line & Beyond Show that (i) Δ DAP Δ EBP (ii) AD = BE DAP EBP BAD ABE (given) EPA DPB (given) So, EPA EPD DPB EPD Or, DPA EPB Answer: In AP=PB (Since P is mid point) So, by ASA rule DAP EBP So, AD=BE A D M B 8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that: (i) Δ AMC Δ BMD (ii) DBC is a right angle. (iii) Δ DBC Δ ACB (iv) CM = 1 AB 2 Answer: In AMC & BMD BM=AM (M is midpoint) DM=CM (given) DMB AMC (opposite angles) So, AMC BMD Hence, DB=AC DBA BAC So, DB||AC (alternate angles are equal) So, BDC ACB = Right Angle) (internal angles are complementary in Case of transversal of parallel lines) DBC & ACB DB=AC (proved earlier) BC=BC (Common side) BDC ACB (proved earlier) So, DBC ACB So, AB=DC So, AM=BM=CM=DM www.excellup.com ©2009 send your queries to [email protected] C Finish Line & Beyond So, CM= 1 AB 2 EXERCISE 2 1. In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that : (i) OB = OC (ii) AO bisects A A Answer: In OBC OBC OAC (they are half of angles B & C) O So, OB=OC ( Sides opposite to equal angles) In AOB & AOC AB=AC (given) OB=OC (proved earlier) C B ABO ACO (they are half of angles B & C) So, AOB AOC (SAS Rule) So, BAO CAO It means that AO bisects A A 2. In Δ ABC, AD is the perpendicular bisector of BC. Show that Δ ABC is an isosceles triangle in which AB = AC. Answer: In ABD & ACD AD=AD (common side) BD=CD (given) B C D ADB ADC (right angle) So, ABD ACD A So, AB=AC, which proves that ABC is isosceles 3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal. Answer: In ABE & ACF AB=AC (given) F B BAE CAF (common to both triangles) CFA BEA (right angles) www.excellup.com ©2009 send your queries to [email protected] E C Finish Line & Beyond So, ABE ACF ( ASA Rule ) So, BE=CF 4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that (i) Δ ABE Δ ACF (ii) AB = AC, i.e., ABC is an isosceles triangle. Answer: This can be solved like previous question. 5. ABC and DBC are two isosceles triangles on the same base BC. Show that = ACD. ABD A B C D D Answer: ABC ACB DBC DCB So, ABC DBC ACB DCB ABD ACD 6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that BCD is a right angle. A C Answer: In ADC & ABC AD=AB AC=AC ACB ABC ACD ADC In ΔABC, ACB ABC CAB 180 www.excellup.com ©2009 send your queries to [email protected] B Finish Line & Beyond CAB 180 2ACB ------------------------------ (1) Similarly in ΔADC, DAC 180 2ACD ------------(2) As BD is a straight line, so CAB DAC 180 So, adding equations (1)&(2) we get 180 360 2ACB 2ACD 180 360 2(ACB ACD ) 2(ACB ACD ) 180 ACB ACD BCD 90 7. ABC is a right angled triangle in which and AB = AC. Find B and C. A = 90° Answer: If AB= AC then angles opposite to these sides will be equal. As you know the sum of all angles of a triangle is equal to 180°, So, A+ B+ C=180° Or, 90°+ B+ C=180° Or, B+ C=180°-90°=90° Or, B= C=90° 8. Show that the angles of an equilateral triangle are 60° each. Answer: As angles opposite to equal sides of a triangle are always equal. So, in case of equilateral triangle all angles will be equal. So they will measure one third of 180°, which is equal to 60° EXERCISE 3 1. Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC . If AD is extended to intersect BC at P, show that (i) Δ ABD Δ ACD (ii) Δ ABP Δ ACP (iii) AP bisects A as well as D. (iv) AP is the perpendicular bisector of BC. Answer: In ABD & ACD AB=AC BD=CD AD=AD So, Δ ABD Δ ACD (SSS Rule) A D C B In ABP & ACP AB=AC AP=AP www.excellup.com ©2009 send your queries to [email protected] P Finish Line & Beyond ABP ACP (Angle opposite to equal sides) So, Δ ABP Δ ACP (SAS Rule) Since Δ ABP Δ ACP So, BAP CAP So, AP is bisecting BAC Similarly BDP & CDP can be proved to be A P Congruent and as a result it can be proved that AP is bisecting BDC 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects A. B M C Q R N Answer: This can be solved like previous question. 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that: (i) Δ ABM Δ PQN (ii) Δ ABC Δ PQR A Answer: In ABM & PQN AB=PQ AM=PN BM=QN (median bisects the base) ABM PQN In ABC & PQR F E So, AB=PQ BC=QR AC=PR (Equal medians means third side will be equal) C B So, ABC PQR 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles. A Answer: In AEB & AFC BE=CF (Perpendicular) AB=BC (Hypotenuse) So, AEB AFC 5. ABC is an isosceles triangle with AB = AC. Draw AD BC to show that B = C. Answer: After drawing AD In ADC & ADB BC B AC=AB www.excellup.com ©2009 send your queries to [email protected] D C Finish Line & Beyond AD=AD ADC ADB So, ADC ADB So, ACD ABC EXERCISE 4 1. Show that in a right angled triangle, the hypotenuse is the longest side. A Answer: In a right angled triangle, the angle opposite To the hypotenuse is 90°, while other two angles are Always less than 90°. As you know that the side opposite to the largest angle is always the largest in a triangle. 2. In the given triangle sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB. B C Q P Answer: ABC 180 PBC ACB 180 OCB Since PBC OCB So, ABC ACB As you know side opposite to the larger angle is larger than the side opposite to the smaller angle. Hence, AC>AB 3. In the given figure Show that AD < BC. D B B < A and C < D. O A C Answer: AO<BO (Side opposite to smaller angle) DO<CO (Side opposite to smaller angle) So, AO+DO<BO+CO Or, AD<BC D 4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that A > C and B > D. Answer: Let us draw two diagonals BD and AC as shown in the figure. In ΔABD Sides AB<AD<BD So, ADB ABD ----------------- (1) Angle opposite to smaller side is smaller In ΔBCD Sides BC<DC<BD So, BDC CBD ---------------- (2) A B www.excellup.com ©2009 send your queries to [email protected] C Finish Line & Beyond Adding equation (1) & (2) ADB BDC ABD CBD ADC ABC Similarly in ABC BAC ACB ------------------ (3) In ADC DAC DCA ----------------- (4) Adding equations (3) & (4) BAC DAC ACB DCA BAD BCD 5. In following figure, PR > PQ and PS bisects QPR. Prove that PSR > PSQ. P 4 5 3 1 2 Q S R Answer: For convenience let us name these angles as follows: PQR 1 PRQ 2 QPR 3 QPS 4 RPS 5 PSQ 6 PSR 7 Since, PR>PQ , so In 1 2 PQS www.excellup.com ©2009 send your queries to [email protected] Finish Line & Beyond 1 4 6 180 In PRS 2 5 7 180 In both these triangles 4 5 1 2 So, for making the sum total equal to 180° the following will always be true: 6 7 www.excellup.com ©2009 send your queries to [email protected]