Question

if the bisector of an angle of triangle bisects the opposite side prove that the triangle is isosceles

Answer 1

In a ∆ABC,Consider AD be the bisector of ∠A then BD = CD.

To prove that ∆ABC is an isosceles triangle i.e. AB = AC. Draw a line from C i.e CE parallel AD .

BA is extended then they meet at E.

Given that ∠BAD = ∠CAD    .............. (i)
 
CE || AD
 
∴ ∠BAD = ∠AEC  (Corresponding angles)  ................ (ii)
 
And ∠CAD = ∠ACE (Alternate interior angles)  .................. (iii)
 
From (i), (ii) and (iii) we obtain
 
∠ACE = ∠AEC
 
In ∆ACE, ∠ACE = ∠AEC
 
∴ AE = AC (Sides opposite to angles are equal)  ................ (iv)
 
In  a ∆BEC, AD||CE and D is the mid-point of BC by converse of mid-point theorem A is the mid-point of BE.
 
∴ AB = AE
 
⇒ AB = AC  [equ (iv)]
 
In a ∆ABC, AB = AC
 
∴ ∆ABC is an isosceles triangle.

Answer 2

It is simple if you know this standard property:

In ABC, the angle bissector in A splits BC in to segments BH and HC proportional to AB and AC.

So if BH = HC then AB = AC.

The property itself follows from Thales.