<s>[INST] What's the measure of height relative to $BC$ in the triangle below?. For reference:In the isosceles triangle $ABC$ ($AB = BC$) is marked
the interior point O whose distances to sides $AB, BC and AC$ are $2, 3$ and $5$ respectively,
If the sum of the height measurements of triangle $ABC$ is $30$, calculate the
measure of height relative to $BC$.
My progress.. (drawing without scale)

I try :
$BF = h_1, CL = HK = h_2$
$HM \parallel FB \implies\\
\angle IMO \sim IBF:\\
\triangle AMH \sim ABF\\
\triangle OIM \sim \triangle AFB \implies\\
\frac{IM}{FB}=\frac{IO}{AH}=\frac{OM}{AB} \rightarrow \frac{IM}{h_1}=\frac{2}{AF}=\frac{OM}{AB}\\
\triangle ION \sim \triangle KCA \sim \triangle LAC$
but I can't finish...
 [/INST] If altitude from $A$ and $C$ to the opposite sides is $h_1$ and from $B$, it is $h_2$, then
$2 h_1 + h_2 = 30 \tag1$
If $A$ is the area of $\triangle ABC$,
$ \displaystyle a = c = \frac{2A}{h_1}, b = \frac{2A}{h_2}$
Now note that $A = Area ~ \triangle OAB + Area ~ \triangle OBC + Area ~ \triangle OAC$
$ \displaystyle A = \frac{1}{2} \left[ 3 a + 2c + 5 b \right] = 5A \left(\frac{1}{h_1} + \frac{1}{h_2}\right) $
$ \implies \displaystyle \frac{1}{h_1} + \frac{1}{h_2} = \frac{1}{5} \tag2$
Using $(1)$ and $(2)$, can you solve for $h_1$?
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