If the angle bisectors of one …

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

BD = CD [Given]

AD = ED [By construction]

and $\angle$ADB = $\angle$EDC [Vertically opposite angles]

Therefore, $\Delta ABD=\Delta ECD$ [SAS congruence criterion]

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