Triangle $ABC$ is an isosceles triangle with $AC = BC$. Point $D$ lies on the extension of side $AC$ past point $C$ such that $CD = AC$. Prove that triangle $ABD$ is a right triangle.

1. Problem Description

Triangle $ABC$ is an isosceles triangle with $AC = BC$. Point $D$ lies on the extension of side $AC$ past point $C$ such that $CD = AC$. Prove that triangle $ABD$ is a right triangle.

2. Solution Steps

Let $\angle BAC = \angle ABC = \alpha$.

Since $AC=BC$, we have that triangle $ABC$ is an isosceles triangle. Therefore, $\angle BAC = \angle ABC = \alpha$.

The sum of angles in a triangle is $180^\circ$, so $\angle ACB = 180^\circ - 2\alpha$.

Since point $D$ lies on the extension of side $AC$ past point $C$, $\angle BCD$ and $\angle ACB$ are supplementary angles.

Thus, $\angle BCD = 180^\circ - \angle ACB = 180^\circ - (180^\circ - 2\alpha) = 2\alpha$.

We are given that $CD=AC$ and $AC=BC$. Therefore, $CD=BC$, which means triangle $BCD$ is isosceles with $CD = BC$.

Then $\angle CDB = \angle CBD$. The sum of angles in triangle $BCD$ is $180^\circ$.

$\angle CDB + \angle CBD + \angle BCD = 180^\circ$

$2\angle CDB + 2\alpha = 180^\circ$

$2\angle CDB = 180^\circ - 2\alpha$

$\angle CDB = 90^\circ - \alpha$

Now we need to find $\angle ABD$. We have $\angle ABD = \angle ABC + \angle CBD = \alpha + \angle CBD$.

Since $\angle CBD = \angle CDB$, $\angle CBD = 90^\circ - \alpha$.

Thus $\angle ABD = \alpha + 90^\circ - \alpha = 90^\circ$.

Since $\angle ABD = 90^\circ$, triangle $ABD$ is a right triangle with the right angle at vertex $B$.

3. Final Answer

Triangle $ABD$ is a right triangle.