Question on isosceles triangle geometry

Question

Find the length of the base of an isosceles triangle ABC with base AC, if it is known that, the length of its heights are AN and BM which equal to n and m respectively.

Tried to do this question. I still can't see the trick of answering here. Somebody help me here please?

Answer 1

Hint:

If $S$ is the area of the triangle then $$2S=m\cdot AC=n\cdot AB$$

Also we have $$AB^2=m^2+\left(\frac{AC}2\right)^2$$

So you have a system. Solve it for $AB$ and $AC$.

Answer 2

Let $NC=x$ and $CM=y$.

First, by Pythagorus theorem, you have $$y^2+(2x)^2=m^2.$$ On the other hand, since $\triangle ANC\sim\triangle BMC$, you have $$\frac{m}{n}=\frac{y}{x}.$$ Solving this system will yield the solution. ($BC=2x$)