In a △ABC, AB = AC. If the bisectors of ∠B and ∠C meet AC and AB at points D and E respectively, show that :

(i) △DBC ≅ △ECB

(ii) BD = CE

In △ABC,

AB = AC

⇒ ∠ABC = ∠ACB (Angles opposite to equal sides in a triangle are equal)

Given,

∠ABD = ∠DBC (DB is bisector of ∠B) ….(1)

∠ACE = ∠ECB (CE is bisector of ∠C) ….(2)

Since, ∠ABC = ∠ACB, from eq.(1) and (2), we have:

⇒ ∠ABD = ∠DBC = ∠ACE = ∠ECB

(i) In △ECB and △DBC,

⇒ BC = BC (Common side)

⇒ ∠ECB = ∠DBC (Proved above)

⇒ ∠EBC = ∠DCB (As, ∠ABC = ∠ACB)

∴ △ECB ≅ △DBC (By A.S.A. axiom)

Hence, proved that △ECB ≅ △DBC.

(ii) Since, △ECB ≅ △DBC

BD = CE (Corresponding parts of congruent triangles are equal)

Hence, proved that BD = CE.