# If the bisector of angle of a triangle bisects the opposite side, prove that the triangle is isosceles.

Consider the ∆ABC, let AD be the bisector of ∠A and BD = CD. It is required to prove ∆ABC is an isosceles triangle i.e. AB = AC. For this draw a line from C parallel AD and extend BA. Let they meet at E.

It is given that

∠BAD = ∠CAD ... (1)

CE||AD

∴ ∠BAD = ∠AEC (Corresponding angles) ... (2)

And ∠CAD = ∠ACE (Alternate interior angles) ... (3)

From (1), (2) and (3)

∠ACE = ∠AEC

In ∆ACE, ∠ACE = ∠AEC

∴ AE = OAC (Sides opposite to equal angles) ... (4)

In ∆BEC, AD||CE and D is the mid-point of BC using converse of mid-point theorem A is the mid-point of BE.

∴ AB = AE

⇒ AB = AC [Using (4)]

In ∆ABC, AB = AC

∴ ∆ABC is an isosceles triangle.

Hence, proved

Regards

**
**