# 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.

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 ?

• "Angles opposite equal sides are equal" (aka, the Isosceles Triangle Theorem) applies when the angles and sides are parts of a single triangle. You've applied this to angles and sides in different triangles ($\triangle ABD$ and $\triangle ACD$). ... To see the flaw another way, redraw your image so that $\overline{BC}$ is slanted instead of horizontal.
– Blue
Aug 28, 2019 at 6:14
• Yes, I understand it now. Thank you for clearing my confusion. Aug 28, 2019 at 6:19
• Fun fact: It is in general true that the ratio of the two parts of the opposite side ($BD/CD$) is equal to the ratio of the two adjacent sides ($AB/AC$). In this case, the ratio happens to be 1. Aug 28, 2019 at 6:28

Your proof is wrong because you did not prove that $$\Delta ABD\cong\Delta ACD.$$
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?