Find The Angle

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Let $BC$ meet $SP$ and $XZ$ at points $M$ and $N$ respectively. Note that $SP^2= SC^2= SA \times SB$ by Power of a Point, so $SP$ is tangent to the circumcircle of $\triangle ABP.$ It follows that $\angle MPB = \angle BAP.$ Also, being angles in the same segment, $\angle BAP= \angle BXZ.$ Therefore triangles $\triangle BMP$ and $\triangle BXN$ are similar, which gives $\angle BMP = \angle MNX.$ This implies $XZ \parallel SP,$ so $\angle SZX = 180^{\circ} - \angle PSZ= \boxed{120^{\circ}}.$

Note: The condition $AB<AC$ makes the configuration similar to the diagram we used. If we had $AB > AC,$ $S$ would be on the other side of $P.$ If we had $AB= AC,$ the tangent through $C$ would be parallel to $AB$ and $S$ would be at infinity.

M=BC x PS, $SC^{2}=SP^{2}=SB \times SA \Rightarrow \frac{SB}{SP}=\frac{SP}{SA}$ $\Rightarrow \triangle{SBP}$ is isoforms with $\triangle{SAP}$ $\Rightarrow \angle{BAP}=\angle{BPS}$ $\Rightarrow \angle{BXZ}=\angle{BPS} \Rightarrow SP \| XZ$ $\Rightarrow \angle{AZX}=60^{o}$ $\Rightarrow \angle{SZX}=\angle{SZA}+60^{o}=\boxed{120^{o}}$