Given: x=y and AB=CB

To prove: AE = CD

Proof:

In ∆ABE, we have,

∠AEC = ∠EBA + ∠BAE …Exterior angle theorem

y° = ∠EBA + ∠BAE

Now in ∆BCD, we have,

x° = ∠CBA + ∠BCD

Since, given that,

x = y ,

∠CBA + ∠BCD = ∠EBA + ∠BAE

∴ ∠BCD = ∠BAE … as ∠CBA and ∠EBA and same angles.

Hence in ∆BCD and ∆BAE,

∠B = ∠B

BC = AB …given

∠BCD = ∠BAE

Thus by ASA property of congruence, we have,

∆BCD ≅ ∆BAE

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ CD = AE