For reference: If $ABCD$ is a square and $BC$ is a diameter.
Show that $x = \sqrt{ab}$.
Progress: Let $h$ = height $\triangle TPQ$
$$\triangle TPQ \sim \triangle TAD$$$$ \frac{h+a+b+x}{h} = \frac {a+b+x}{x} \iff \frac{a+b + x}{h} = \frac{a+b}x$$
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Setting $l=a+b+x$ your result can be written as: $$ {l\over h}={l-x\over x}, \quad\text{that is:}\quad h={lx\over l-x}. $$ On the other hand, in right triangle $BCT$ we have: $$ h^2=(a+x_1)(b+x_2), $$ where $x_1$ and $x_2$ are the projections of $PT$ and $QT$ on $BC$. But $x_1=ah/l$, $x_2=bh/l$, which inserted into previous formula give: $$ h^2=ab\left(1+{h\over l}\right)^2. $$ Substituting here $h={lx\over l-x}$ we get $x^2=ab$.