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


Here is my solution...



To Prove - Triangle ABC is isosceles or AB = AC.


$BD = CD$ (Given)
$\angle BAD = \angle CAD$ (Given)
$\angle ABD = \angle ACD$ ($AD$ is a common side, angles opposite equal sides are equal)
$\triangle ABD$ and $\triangle ACD$ are congruent as per AAS postulate. 
And therefore, $AB = AC$.


Is this a right answer or am I wrong somewhere ? I've seen solutions for this question but all of them have solved through constructions. I feel this is a shorter and logical way. Am I right or wrong in this approach ?
Your proof is wrong because you did not prove that $\Delta ABD\cong\Delta ACD.$

The hint:

Let $E$ be placed on the line $AD$ such that $D$ is a mid-point of $AE$.

Thus, $$\Delta ADC\cong\Delta EDB,$$ $$\measuredangle BAD=\measuredangle BED.$$
Can you take it from this?
Your answer is wrong as what you were asked to prove, the same you are taking as a statement which is actually not allowed. 
To prove it you have to draw a line from C which is parallel to AD and meeting BA at E. Now prove it with the help of converse of MPT.