It is known that square ABCD is inscribed at O, and point P is a point on inferior arc ad. it connects AP, BP, CP and (AP + CP) / BP (important process)

It is known that square ABCD is inscribed at O, and point P is a point on inferior arc ad. it connects AP, BP, CP and (AP + CP) / BP (important process)

Let angle PAB = x, then angle PBC = 90 degrees - x, because angle PAB + angle PDC = 1 / 2, angle AOD = 1 / 2 * 90 = 45 degrees leads to angle PDC = 45 degrees - angle PAB = 45 degrees - x, so angle PCB = 90 degrees - angle PDC = 90 degrees - 45 degrees + x = 45 degrees + X is obtained by PA / sin (angle PAB) = PC / sin (angle PBC) = BP / sin (angle PCB) = 2R [R is the radius of the outer circle]