In the given figure, ∠ABD = ∠EBC, BD = BC and ∠ACB = ∠EDB. Prove that AB = BE.

Given,

∠ACB = ∠EDB and ∠ABD = ∠EBC

From figure,

⇒ ∠EBD = ∠ABE + ∠ABD …..(1)

Also,

⇒ ∠ABC = ∠ABE + ∠EBC

⇒ ∠ABC = ∠ABE + ∠ABD …..(2)

From eq.(1) and (2), we have :

⇒ ∠ABC = ∠EBD

In △EDB and △ACB,

⇒ DB = BC [Given]

⇒ ∠ABC = ∠EBD [Proved above]

⇒ ∠ACB = ∠EDB [Given]

∴ △EDB ≅ △ACB (By A.S.A. axiom)

⇒ AB = BE [Corresponding parts of congruent triangles are equal.]

Hence, proved that AB = BE.