Problem I'm trying to solve:
Given triangle $ABC$ inscribed in a circle a shown. The altitudes of the triangle meet at point $P$. $AP=21\text{cm}$ and $BC=20\text{cm}$. What is the diameter of the circle?
My Attempt
Let the point where the altitude of $AB$ meets $AB$ be $Q$,the point where the altitude of $AC$ meets $AC$ be $R$ and the point where the altitude of $BC$ meets $BC$ be $S$. Then let $PQ$ be $21x$. From the similarity of $\triangle APQ$ and $\triangle BCQ$, $BQ=20x$ and by the pythagorean theorem $BP=29x$. It's relatively easy to see that $\triangle AQP \sim \triangle ABS \sim \triangle PCS \sim \triangle BCQ$ and pretty much the same for most other triangles that can be found in the circle however I'm lost as to where to go from there since everything else just seems to give lengths of triangle sides in terms of $x$.
I've also tried drawing radii and diameters through the vertices of the triangle in order to determine some possible relation between the angles, sides and the circle's radius but this doesn't seem to get anywhere either.
My original idea was to determine the area of triangle and the sides of the triangle from which the radius readily follows but I just seem to have too little information to move forward with anything? I don't think there's a way to apply Ptolemy's theorem or any other theorem here so I just simply don't know how to move forward.
(Similar to the question attempt,) Consider the similar triangles $\triangle AQP\sim \triangle CQB$,
$$\begin{align*} \frac{AQ}{QC} &= \frac{21}{20}\\ &= \cot\angle QAC \end{align*}$$
Then what follows is essentially the same as what the sine law describes, applying on $\triangle ABC$:
$$\begin{align*} \text{Circumdiameter} &= \frac{BC}{\sin\angle QAC}\\ &= {20}\sqrt{1+\cot^2\angle QAC}\\ &= 29\text{ cm} \end{align*}$$
Construct $A'$ on the same side of $BC$ as $A$, where $A'B \perp BC$ and $A'B = 21\text{ cm}$. (Whether $A'$ is on the same circumcircle is to be proven.)
So these right-angled triangles are similar: $\triangle A'BC \sim \triangle AQC$ (SAS).
The corresponding angles $\angle BA'C = \angle QAC = \angle BAC$ (also equal to $\cot^{-1}\frac{21}{20}$), so $A,A',B,C$ are concyclic.
$A'B \perp BC$ by construction, and the diameter of their circumcircle is
$$A'C = \sqrt{20^2+21^2} = 29\text{ cm}$$