Question

In the following diagram; AD = AB and AE bisect angle A.

Prove that:
(i) BE = DE
(ii) ∠ABD > ∠C

Answer 1

Const: Join ED.

In ΔAOB and ΔAOD,
AB = AD                 ...[ Given ]
AO = AO                ....[ Common ]
∠BAO = ∠DAO      ....[ AO is bisector of A ]
∴ ΔAOB ≅ ΔAOD  ....[ SAS criterion ]

Hence,
BO = OD                                                  …(i)[ c.p.c.t. ]
∠AOB = ∠AOD                                        …(ii)[ c.p.c.t. ]
∠ABO = ∠ADO ⇒ ∠ABD = ∠ADB            …(iii)[ c.p.c.t. ]

Now,
∠AOB = ∠DOE                ...[Vertically opposite angles]
∠AOD = ∠BOE               ...[Vertically opposite angles]
∠BOE = ∠DOE               …(iv)[ From (ii) ]

(i) In ΔBOE and ΔDOE,
BO = OD                               ...[ From (i) ]
OE = OE                               ...[ Common ]
∠BOE = ∠DOE                     ...[ From (iv) ][ SAS criterion ]
Hence, BE = DE                    ...[ c.p.c.t. ]

(ii) In BCD,
∠ADB = ∠C + ∠CBD        ...[ Ext. angle = sum of opp. int. angles ]
⇒ ∠ADB > ∠C
⇒ ∠ABD > ∠C                     ...[ From (iii) ]