In the given figure, AD = AB and AE bisects ∠A. Prove that:

(i) BE = ED

(ii) ∠ABD > ∠BCA

In △ABE and △ADE,

⇒ AE = AE (Common side)

⇒ AB = AD (Given)

⇒ ∠BAE = ∠DAE (∵ AE bisects ∠A)

∴ △ABE ≅ △ADE (By S.A.S axiom)

(i) Since, △ABE ≅ △ADE

We know that,

Corresponding parts of congruent triangle are equal.

∴ BE = ED

Hence, proved that BE = ED.

(ii) Since, △ABE ≅ △ADE

We know that,

Corresponding parts of congruent triangle are equal.

∴ ∠ABE = ∠ADE

⇒ ∠ABD = ∠ADB

From figure,

⇒ ∠BDA > ∠BCA (∵ Exterior angle is greater than interior opposite angle)

⇒ ∠ABD > ∠BCA

Hence, proved that ∠ABD > ∠BCA.