How to prove $C'$ is reflection of $C$ when we know $\triangle ABC\cong \triangle ABC'$?

Question

We know $\triangle ABC\cong \triangle ABC'$ (the triangles are congruent). We may assume $C\neq C'$. How do I prove $C'$ is the reflection of $C$ in the line $AB$?
I know I first need to prove the fact that the orthogonal projections of $C$ and $C'$ on the line $AB$ are the same, but I really don't know how to start.

Answer 1

ABC, ABC' are congruent mounted on the same side AB. Since CC' should make equal angles with AB and since thay are adjacent, ( total angle $180^{\circ}$) each angle is $90^{\circ}$, C reflects to C'.

ACBC' is a kite.