In triangle ABC, angle A = angle C, and AB = AC. D is a point on side AC such that BD is perpendicular to AC. Prove that triangle ABD is isosceles.

To prove that triangle ABD is isosceles, we need to show that AD = BD.

Given that triangle ABC is isosceles (since AB = AC and angle A = angle C), we know that the base angles of an isosceles triangle are equal. Therefore, angle ABC = angle ACB.

Since BD is perpendicular to AC, angle ABD = angle ABC = 90 degrees.

Similarly, angle ADB = angle ACB = 90 degrees.

Now, we have two triangles ABD and ADB with:

1. A common side AD.
2. Angle ABD = angle ADB = 90 degrees.
3. Angle BAD = angle BDA (since they are both equal to angle BAC).

By the Angle-Angle-Side (AAS) criterion, we can say that triangle ABD is congruent to triangle ADB.

Since corresponding parts of congruent triangles are equal, we have AD = BD.

Therefore, triangle ABD is isosceles.