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 ?