Find the area of $ABCD$ 

Find the area of  $ABCD$ if $PSC$ is a semicircle.

I have used the tangent properties of circles. I also assumed many variables to get a relationship between product of sides of the rectangle $ABCD$ but eventually and unfortunately, I proved $$1=1$$
Any help is greatly appreciated.
 A: We can see in triangles PQC and QBC that
$∠QPC = ∠BQC,∠QCP$ is common.
$∠PQC = ∠QBC = 90$
Therefore both triangles are similar.
$\frac{QC}{PC}=\frac{BC}{QC}$,
$PC∗BC=10^{2}=100$
Now We can say that one side CD is equal to the radius and so diameter $PC=2∗radius = 2*CD$
Area=$BC*CD = \frac{BC∗PC}{2}=\frac{100}{2}=50$
A: Let $E$ be the projection of $D$ on $CQ$. Move triangle $BCQ$ by $\vec{BA}$. The question boils down to finding the area of a parallelogram whose area is equal to $CQ\cdot DE =10\cdot DE$.
Let $M$ be the midpoint of $PC$, and $R$ be the midpoint of $CQ$. Can you prove that triangles $CDE$ and $MCN$ are congruent? This will lead to $DE=5$, hence the area equals $50$.
A: Lets call lengths:
$\overline {BC} = x\\
\overline {PB} = y\\
\overline {BQ} = z$
$\frac {x+y}2$ is the raduis of the circle.
$\frac {x^2+xy}{2}$  is the area of the rectangle.
$x^2+z^2 = 10^2$  by Pythagoras
$xy = z^2$  I not sure the name of the theorem that provideds this result.
$x^2 + xy = 100$
But $x^2 + xy$ is double the area of the rectanlge
$A = \frac {x^2+xy}{2} = 50$
