Question

In triangle ABC, ray BD bisects angleABC and ray,
CE bisects angle ACB. If seg AB =segAC then
prove that ED II BC
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Answer 1

GIVEN: Triangle ABC , AB = AC => < B = <C

BD bisects < B & CE bisects < C

TO PROVE: ED // BC

PROOF: < B = < C ( given)

=> <B/2 = < C/2

=> < ABD = < ACD

In triangle ABD & triangle ACE

< ABD = < ACE ( proved above)

<A = A ( common angle)

=> tri ABD ~ tri ACE ( by AA similarity criterion)

=> AB/AC = AD/AE ( csst)

But AB = AC ( given) …….. (1)

=> AD = AE ………(2)

=> AB - AE = AC - AD

=> BE = DC …….. (3)

By (2) & (3)

ED // BC ( by converse of Basic proportionality theorem)

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