Question

ABCD is a cyclic quadrilateral inscribed in the circle \Gamma with AB as diameter. Let E be the intersection of the diagonals AC and BD. The tangents to \Gamma at the points C,D meet at P. Prove that PC=PE.

Answer 1

Given that AB is the diameter of the circle, we have ∠ACB = 90°, and similarly ∠ADB = 90°.Since ABCD is a cyclic quadrilateral, we have ∠AEB + ∠CED = 180°.Also, since AC and BD are diagonals of the quadrilateral, they intersect at point E.Consider the tangent to the circle at point C. Since PC is also a tangent to the circle, we have ∠CAB = ∠PCB.