Prove that a circle drawn on one of the equal sides of an isosceles triangle as diameter bisects the base of the triangle.

                                                    
In the figure ΔABC is an isosceles such that AB = BC. Circle C(O, r) is drawn on the side AB as diameter and intersecting side AC in D.
To Prove: AD = CD
Construction: Join B to D
Proof: ∠ADB = 90° (angle in a semi-circle)
Now in ΔABD and ΔBDC
AB = BC (given)
BD = BD (common)
∠ADB = ∠BDC (each 90°)
∴ ΔABD ≅ ΔBDC
∴ AD = CD