# What’s the measure of the segment $PQ$ in the figure below

euclidean-geometryplane-geometry

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 $$PN \therefore AM =\frac{R}{2}=OR\\ \triangle BQM:(notable) \implies MQ = \frac{3R\sqrt5}{10}$$

I couldn't relate PQ…

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 $$\triangle PQD$$.