FKMO 2005 Problem 4

In the following, the point of intersection of two lines $g$ and $h$ will be abbreviated as $g\cap h$.

Suppose $ABC$ is a triangle in which $\angle A = 90^{\circ}$ and $\angle B > \angle C$. Let $O$ be the circumcircle of the triangle $ABC$. Let $l_{A}$ and $l_{B}$ be the tangents to the circle $O$ at $A$ and $B$, respectively.

Let $BC \cap l_{A} = S$ and $AC \cap l_{B} = D$. Furthermore, let $AB \cap DS = E$, and let $CE \cap l_{A} = T$. Denote by $P$ the foot of the perpendicular from $E$ on $l_{A}$. Denote by $Q$ the point of intersection of the line $CP$ with the circle $O$ (different from $C$). Denote by $R$ be the point of intersection of the line $QT$ with the circle $O$ (different from $Q$). Finally, define $U = BR \cap l_{A}$. Prove that $$\frac {SU \cdot SP}{TU \cdot TP} = \frac {SA^{2}}{TA^{2}}.$$