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

Here is my solution...

enter image description here

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 ?

  • $\begingroup$ "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. $\endgroup$
    – Blue
    Aug 28, 2019 at 6:14
  • 1
    $\begingroup$ Yes, I understand it now. Thank you for clearing my confusion. $\endgroup$ Aug 28, 2019 at 6:19
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    $\begingroup$ 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. $\endgroup$
    – Arthur
    Aug 28, 2019 at 6:28

1 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?

  • $\begingroup$ Yes, I can prove it from here, thank you ! $\endgroup$ Aug 28, 2019 at 6:17
  • $\begingroup$ @Bhashwar Sengupta You are welcome! $\endgroup$ Aug 28, 2019 at 6:19

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