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…
Here is an approach that avoids trigonometry. In $\triangle ADH$, $\frac{AH}{AD} = \frac{3}{5} \implies AD = R = \frac{5}{3} AH$.
As $AH = AC/2$, $R = \frac{5}{6} AC \tag1$
$H$ is the circumcenter of right triangles $\triangle AFC$ and $\triangle AEC$.
So, $FH = EH = AC/2$
$\angle AHF = 180^\circ - 2 \angle A, \angle EHC = 180^\circ - 2 \angle C$
That leads to $\angle EHF = 180^\circ - 2 \angle B$
As $\triangle EHF$ is isosceles and $M$ is the foot of perpendicular from $H$ to $FE$,
$\angle HFM = \angle HEM = \angle B$ and $FM = ME = 12$.
In $\triangle FHM$, $ \displaystyle \frac{FM}{FH} = \frac{4}{5}$
$FH = 15 = AC/2 \implies AC = 30$
Using $(1)$, $R = 25$
Many thanks to @Ivan for identifying the error in the original image, and here is a solution by geometry as per the request of OP.
Extend $\small PH$ to meet $\small BC$ extend at $\small G$. $\angle CGH=53^\circ$.
$\small \therefore\angle BGP=\angle BAP\implies AGBP\ \text{is cyclic}$
As $\small AB=AP$, $\small \angle ABP=63.5^\circ$, and so is $\small \angle AGP$ because they are in the same segment.
Therefore, $\small \triangle AHG$ is a special right triangle with perpendicular sides in the ratio 1:2.
$\small \therefore HG=6$
Similarly, as $\small \triangle CGH$ is a $\small 3:4:5$ right triangle, $\small HC=8$.
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 $\triangle PQD$.