For reference: In the figure calculate PQ(Answer:$\frac{R\sqrt5}{5}$)

My progress:

$C=DB \cap \circ AB\\

\triangle ACB: AC = R, AB = 2R(notable)\\

\therefore \angle ABC = 30^o, BC = R\sqrt3\\

\triangle BAD: BD = R\sqrt5$

Extend $AD$ to $N (NA = AD)$

Draw $PN \therefore AM =\frac{R}{2}=OR\\

\triangle BQM:(notable) \implies MQ = \frac{3R\sqrt5}{10} $

## Best Answer

Sides of $\triangle DAB$ are in ratio $1:2:\sqrt5$

As $\triangle ARB \sim \triangle DAB$

$AR = \frac{2R}{\sqrt5}, BR = \frac{4R}{\sqrt5} $

As $\angle DAB = 90^\circ, $ and $AR \perp BD$, $ \displaystyle AR^2 = BR \cdot RD \implies RQ = RD = \frac{R}{\sqrt5}, BQ = \frac{3R}{\sqrt5}$.

So, $ \displaystyle AT = CT = \frac{R}{2}$ and that leads to $QT = \frac{3R}{2\sqrt5}, QS = \frac{R}{\sqrt5} ~ , ~ $ where $S$ is the foot of the perp from $C$ to $QT$.

By intersecting chords theorem,

$ \displaystyle BQ \cdot RQ = \frac{3R^2}{5} = PQ \cdot (2 PS - PQ) = x \left(x + \frac{2R}{\sqrt5}\right)$

Solving, $ \displaystyle x = \frac{R}{\sqrt5}$

Alternatively, once you show that $CT = \frac{R}{2}$, you know that perp from $P$ to $AB$ is $R$ (using similarity of triangle to $\triangle DAB$). As the perp is $R$, it must be $CP$ that is perp $AB$. That leads to an alternate solution using Pythagoras in $\triangle CPT$ or $\triangle PQD$.