What's the measure of the circumradius of triangle ABC? For reference (exact copy of the question):

In the acute triangle $ABC$, the distance between the feet of the relative heights to sides $AB$ and $BC$ is $24$. Calculate the measure of the circumradius of triangle $ABC$. $\angle B = 37^\circ$
(Answer:$25$}

My progress:
My figure and the relationships I found

I tried to draw $DH\perp AC$ in $H$ $\implies \triangle DCH$ is notable ($37^\circ:53^\circ$) therefore sides = $3k:4k:5k$
$FE$ is a right triangle cevian...but I can't see where it will go into the solution.
 A: 
Let $BG$ be the altitude from $B$ to $AC$. $\triangle EFG$ is orthic triangle.
A property of orthic triangle: Each line from the circumcenter of the parent triangle to a vertex is always perpendicular to the corresponding side of the orthic triangle.
As $D$ is the circumcenter of $\triangle ABC$, $ ~ CD \perp EG ~ $ and $AD \perp FG$.
So, $\angle EGF = 180^\circ - \angle ADC = 106^\circ$
If $R_O$ is the circumradius of $\triangle EGF$,
using extended sine law in $\triangle EGF, EF = 24 = 2 R_O \sin \angle EGF$
$\implies \displaystyle R_O = \frac{12}{\sin 106^\circ} \tag1$
$ \sin 106^\circ = 2 \sin 53^\circ \cos 53^\circ = \displaystyle 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$
So from $(1)$, $ \displaystyle R_O = \frac{25}{2}$
Now we use another property of orthic triangle: Circumradius of orthic triangle is half the circumradius of the parent triangle.
$\therefore R = 2 R_O = 25$
A: 
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 $(1)$, $R = 25$
A: 

Property: In a triangle $ABC$, the side lengths of the orthic triangle are given by
$\begin{align}a'&=a\cos A\\b'&=b\cos B\\c'&=c\cos C\end{align}$


Accordingly, in your image,



$EF=AC\cos B$.
As we know $EF=24$ (given) and $\cos B=\cos 37^\circ=\frac45$, this implies $AC=30$.
Therefore $CH=\displaystyle{\frac{AC}2}=15$.
From the special right triangle $DCH$, we find $DC=\frac53CH=25$. $$\therefore R=25$$
A: Let altitudes $AE$ and $CF$ intersect at orthocenter $J$.
Since $\angle BEJ + \angle JFB = 90^\circ + 90^\circ = 180^\circ$, $BFJE$ is a cyclic quadrilateral with diameter $BJ$, by sine rule in $\triangle BFE$,
$$ \frac{EF}{\sin B} = BJ \tag{1}$$
Denote circumradius of $\triangle ABC$ by $R$. Then in terms of $R$ we have formula for $BJ$ as
$$BJ = 2DH = 2R\cos B \tag{2}$$
Combining both equations, one gets $\boxed{R=25}$.
