Orthocenter of a Triangle

Question

Given an Isosceles triangle where in the altitude to the base is divided into lengths 7 and 9 by the orthocenter, what are the sides of the triangle.

I solved this using coordinate geometry using perpendicular lines for the slopes and such. the answer is that the legs are 20 and the base is 24.

I want to know another approach where in we do not use coordinate geometry, I have tried using similar triangles, using the orthocenter but I am stuck, I am thinking that we would use a property (?) of an orthocenter that I have ignored

from ITMO national parallel cluster, Philippines (does not affect the actual ITMO) awards have already been given out.

Answer 1

Hint: Consider $A,B,C$ to be vertices $E,F,G$ the respective perpendicular points to the vertices and $O$ the orthocenter. Notice $AOF$ is similar to $ACE$ and using this information you can find the lengths

Answer 2

Let $|AB|=c$, $\angle C_hCA=\tfrac\gamma2$.

$\triangle C_hCA$ is similar to $\triangle C_hAO_h$,

\begin{align} \angle C_hCA&=\angle C_hAO_h=\tfrac\gamma2 ,\\ \triangle AC_hC:\quad \tan\tfrac\gamma2&=\frac{c/2}{16}=\frac{c}{32} ,\\ \triangle O_hC_hA:\quad \tan\tfrac\gamma2&=\frac{9}{c/2}=\frac{18}{c} ,\\ \frac{18}c&=\frac{c}{32} ,\\ c&=24 . \end{align}