If altitude from one vertex of …

Given: In ∆ ABC, AD is the angular bisector of {tex}\angle{/tex}A which meets BC in D such that BD = DC........(a)
To prove:  ∆ABC is an isosceles.
Construction: Produce AD to E such that AD = DE and then join CE.
Proof: In ∆ ABD and ∆ ECD, we have
BD = CD [ from (a) ]
AD = ED (By construction)
and {tex}\angle{/tex}ADB = {tex}\angle{/tex}EDC (Vertically opposite angles)
Therefore, ∆ ABD {tex}\cong{/tex} ∆ ECD (SAS congruency criterion of triangle)
So, AB = EC (CPCT) ...(i)
and {tex}\angle{/tex}BAD = {tex}\angle{/tex}CED (CPCT) ...(ii)
Also, {tex}\angle{/tex}BAD = {tex}\angle{/tex}CAD (Given, BD is angular bisector of {tex}\angle A{/tex} ).....(iii)
Therefore, from (ii) and (iii)
{tex}\angle{/tex}CAD = {tex}\angle{/tex}CED
So, AC = EC (Sides opposite to equal angles are equal ) ...(iv)

From (i) and (iv), we get
AB = AC.

Therefore, ∆ ABC is isosceles. 

Hence, Proved.