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

Question

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.

1. $BD = CD$ (Given)
2. $\angle BAD = \angle CAD$ (Given)
3. $\angle ABD = \angle ACD$ ($AD$ is a common side, angles opposite equal sides are equal)
4. $\triangle ABD$ and $\triangle ACD$ are congruent as per AAS postulate.
5. 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 ?

Answer 1

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?