AB and CD are the diameters of a circle which intersect at P. Join AC, CB, BD and DA. If \angle PAD=60^\circ, then what is \angle BPD equal to?

Explanation

P is the center of the circle, so PA=PD (radii). Thus, \Delta PAD is isosceles. Since \angle PAD = 60^\circ, it becomes an equilateral triangle, and \angle APD = 60^\circ. Since AB is a straight line, \angle BPD = 180^\circ - \angle APD = 180^\circ - 60^\circ = 120^\circ.

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