Let $ABC$ be a right angled triangle with hypotenuse $AB$ and $P$ be a point on the shorter arc $AC$ of the circumcircle of triangle $ABC$. The line, perpendicuar to $CP$ and passing through $C$, intersects $AP$, $BP$ at points $K$ and $L$ respectively. Prove that the ratio of area of triangles $BKL$ and $ACP$ is independent of the position of point $P$.
We will denote the area of a triangle $XYZ$ by $[XYZ]$.First it is easy to observe that $\Delta PCL \sim \Delta KPL \sim \Delta ACB$ .Hence $\frac{KL}{PL}=\frac{AB}{CB}$ and $\frac{PK}{KC}=\frac{AB}{AC}$ .Also a little angle chasing gives that $\Delta LBC \sim \Delta PAC \Rightarrow \frac{LB}{PA}=\frac{BC}{AC}$
Now
$\frac{[LKB]}{[PAC]}=\frac{[LKB]}{[LKP]}.\frac{[LKP]}{[CKP]}.\frac{[CKP]}{[APC]}=\frac{LB}{LP}.\frac{LK}{KC}.\frac{KP}{PA}=\frac{LB}{PA}.\frac{LK}{LP}.\frac{KP}{KC}=\frac{CB}{CA}.\frac{AB}{BC}.\frac{AB}{AC}=\frac{AB^2}{AC^2}$ which is independent of the choice of $P$