What’s the measure of the circumradius of triangle ABC

euclidean-geometryplane-geometry

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

enter image description here

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.

Best Answer

enter image description here

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 $(1)$, $R = 25$

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