In the following diagram; AD = AB and AE bisects angle A. Prove that :

(i) BE = DE

(ii) ∠ABD > ∠C

Join ED.

In △ AOB and △ AOD,

⇒ AB = AD (Given)

⇒ AO = AO (Common)

⇒ ∠BAO = ∠DAO (AO is the bisector of angle A)

∴ △ AOB ≅ △ AOD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ BO = OD ……….(1) [By C.P.C.T.C.]

⇒ ∠AOB = ∠AOD ………(2) [By C.P.C.T.C.]

⇒ ∠ABO = ∠ADO [By C.P.C.T.C.]

⇒ ∠ABD = ∠ADB ……….(3)

From figure,

⇒ ∠AOB = ∠DOE and ∠AOD = ∠BOE (Vertically opposite angles are equal)

Substituting values of ∠AOB and ∠AOD from above equation in equation (2), we get :

⇒ ∠DOE = ∠BOE ………….(4)

(i) In △ BOE and △ DOE,

⇒ BO = OD [From equation (1)]

⇒ OE = OE [Common side]

⇒ ∠BOE = ∠DOE [From equation (4)]

∴ △ BOE ≅ △ DOE (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ BE = DE.

Hence, proved that BE = DE.

(ii) In triangle BCD,

⇒ ∠ADB = ∠C + ∠CBD (An exterior angle is equal to sum of two opposite interior angles)

⇒ ∠ADB > ∠C

⇒ ∠ABD > ∠C (Since, ∠ABD = ∠ADB)

Hence, proved that ∠ABD > ∠C.