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

Dear Student

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)


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

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

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


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


  • 0
What are you looking for?