If the angle bisectors of one …

Given: In a {tex}\triangle{/tex}ABC, consider AD be the bisector of {tex}\angle{/tex}A  which meets BC in D such that BD = DC.
To Prove: {tex}\triangle{/tex}ABC is an isosceles triangle i.e. AB = AC.
Construction: Produce AD to E such that AD=DE and then join CE.
Proof: In {tex}\Delta ABD \;and \;\Delta ECD,{/tex} we have

BD = CD [Given]

AD = ED [By construction]

and {tex}\angle{/tex}ADB = {tex}\angle{/tex}EDC [Vertically opposite angles]

Therefore, {tex}\Delta ABD=\Delta ECD{/tex} [SAS congruence criterion]

So, AB = EC [Corresponding parts of congruent triangles are congruent]  ---- (i)
and {tex}\angle{/tex}BAD = {tex}\angle{/tex}CED [Corresponding parts of congruent triangles are congruent]   ---- (ii)
Also, {tex}\angle{/tex}BAD = {tex}\angle{/tex}CAD [Given] ---- (iii)
Therefore, From (ii) and (iii) we get
{tex}\angle{/tex}CAD = {tex}\angle{/tex}CED
So, AC = EC [Sides opposite to angles are equal]....... (iv)
From (i) and (iv), we get
AB = AC
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC is an isosceles triangle.