Let triangle ABC acutangle, with m∠ACB≤ m∠ABC. M the midpoint of side BC and P a point over the side MC. Let C1 the circunference with center C. Let C2 the circunference with center B. P is a point of C1 and C2. Let X a point on the opposite semiplane than B respecting with the straight line AP; Let Y the intersection of side XB with C2 and Z the intersection of side XC with C1. Let m\angle PAX \equal{} \alpha and m\angle ABC \equal{} \beta. Find the geometric place of X if it satisfies the following conditions:
(a) \frac {XY}{XZ} \equal{} \frac {XC \plus{} CP}{XB \plus{} BP}
(b) \cos(\alpha) \equal{} AB\cdot \frac {\sin(\beta )}{AP} trigonometrygeometry unsolvedgeometry