What's the measure of the circumradius in triangle below?

Question

For reference: In the right triangle $ABC$ ($\angle B=90^o$) a perpendicular is drawn from vertex $C$ to $BC$ to a point $D$ such that $BD$ intersects $AC$ at $P$. Calculate the circumradius of the triangle $BPC$, if $AB =4; BP=n; PC =m$ and $CD =6$ (Answer: $\frac{5mn}{24}$)

My progress:

$\triangle OFC:\\ OF^2+\frac{m^2}{4} = r^2 \implies AO^2 - AF^2 = r^2 - \frac{m^2}{4}\\ \triangle ABC:\\ 4^2+BC^2 = AC^2 \implies 16 + BC^2 =(AP+m)^2 \\ \triangle BCD: \\BC^2 + CD^2 = BD^2\implies BC^2 = CD^2+(n+PD)^2\\ \triangle OFA:\\ FO^2+(AP+\frac{m}{2})^2 = OA^2$

but I'm not being able to equate ....

Answer 1

If $R$ is the midpoint of $AC$ then $R$ is circumcenter of right triangle $\triangle ABC$. So, $AR = RC = BR$.

Also note that if $S$ is midpoint of $BC$ then $R, S$ and $O$ are collinear.

Next we see that $\angle BOR = \angle APB$ and $\angle BRO = \angle BAP$ and we conclude that $\triangle BOR \sim \triangle BPA$.

So, $ ~ \displaystyle \frac{OB}{BP} = \frac{BR}{4} \implies OB = \frac{n \cdot BR}{4}$

Now to find $BR$, we use $\triangle APB \sim \triangle CPD$.

$\displaystyle \frac{4}{6} = \frac{AP}{PB} \implies AP = \frac{2m}{3} \implies AC = \frac{5m}{3}$

So, $ \displaystyle BR = AR = \frac{5m}{6}$

Answer 2

Draw $PE \perp BC$ with $E$ on $BC$. Then $\triangle BEP \sim \triangle OCF$ since $\angle PBE = \angle FOC$. Hence $$\frac{BP}{PE} = \frac{OC}{FC} \tag{1}$$

$PE$, $AB$ and $CD$ are related as (apply basic proportionality theorem to $\triangle$s $ABC$ and $BCD$) $$\frac{1}{PE}=\frac{1}{AB}+\frac{1}{CD} \quad \text{or,} \quad PE=\frac{AB\cdot CD}{AB+CD}=\frac{12}{5}$$

Thus from ($1$), $$r=OC=\frac{BP\cdot FC}{PE} = \frac{n \times (m/2)}{12/5}=\frac{5mn}{24} $$