Answer :

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

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