Geometry proof problem (high school)

Question

I have an upcoming chapter test and this was one of the practice problems. Can someone guide me?

Given: Isosceles $\triangle ABC$ with $AB$ congruent to $AC$; $AD$ is not a median of $\triangle ABC$. Prove: $AD$ does not bisect $\angle A$.

My idea is to use proof by contradiction but I'm not sure if my thinking is right.

Answer 1

What you can do, and this might be what you meant, is to prove the contrapositive, which is kind of like a proof by contradiction, but is more direct. We take as given the isosceles triangle with $AB \cong AC$, with $D$ on $BC$, prove that

Assume $AD$ bisects $\angle A,$ then prove that AD is a median of $\triangle ABC$.

If you can do this, you will have proven

(I) IF $AD$ bisects $\angle A,$, THEN $AD$ is a median of $\triangle ABC$.

Proving (I) is equivalent to proving, with the initial "givens", that

(II) if $AD$ is NOT a median of $\triangle ABC$, then AD does not bisect $\angle A$.

Do you think you can do this? (Think SAS).

Answer 2

I would prove it using the contrapositive.

You want to prove this: If AD is not a median of $\triangle ABC$ (which is the same as D does not bisect BC), then AD does not bisect $\angle A$.

The contrapositive, which is logically equivalent is this: If AD bisects $\angle A$ then AD is a median of $\triangle ABC$ (which is the same as D bisects BC).

This should be easy to prove (SAS).

Answer 3

If you want to proof by contradiction, then you can do this.

Assume that $AD$ is not median and bisects $\angle BAC$, as it's shown on the picture below

then, according to SAS ($AB = AC$ by statement of the problem, $\angle BAD = \angle DAC$ by assumption and $AD$ is common edge) $\triangle BAD \cong \triangle ADC$ which means $BD = DC$ as well, which in turn means that $D$ is median. Contradiction.