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∆ ABC and ∆ DBC 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 ∆ ABC =~ ∆ ACD |
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Answer» ong>Answer: (i) In △ABD and △ACD, AB=AC ....(since △ABC is isosceles) AD=AD ....(common side) BD=DC ....(since △BDC is isosceles) ΔABD≅ΔACD .....SSS test of congruence, ∴∠BAD=∠CAD i.e. ∠BAP=∠PAC .....[c.a.c.t]......(i) (II) In △ABP and △ACP, AB=AC ...(since △ABC is isosceles) AP=AP ...(common side) ∠BAP=∠PAC ....from (i) △ABP≅△ACP .... SAS test of congruence ∴BP=PC ...[c.s.c.t].....(ii) ∠APB=∠APC ....c.a.c.t. (iii) Since △ABD≅△ACD ∠BAD=∠CAD ....from (i) So, AD BISECTS ∠A i.e. AP bisects∠A.....(iii) In △BDP and △CDP, DP=DP ...common side BP=PC ...from (ii) BD=CD ...(since △BDC is isosceles) △BDP≅△CDP ....SSS test of congruence ∴∠BDP=∠CDP ....c.a.c.t. ∴ DP bisects∠D So, AP bisects ∠D ....(IV) From (iii) and (iv), AP bisects ∠A as well as ∠D. (iv) We know that ∠APB+∠APC=180 ....(angles in linear pair) Also, ∠APB=∠APC ...from (ii) ∴∠APB=∠APC = =90 ° BP=PC and ∠APB=∠APC=90° Hence, AP is perpendicular bisector of BC. . . Hope you Understand , Please like ❤️ Thank you |
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