What is the height of the intersection point when two $\triangle$ are faced. One $\triangle$ has the base 12 and the other 9
What I have tried: The triangles are similar, which means that the sides of the larger is $\dfrac{12}{9}$ or $\dfrac{4}{3}$ times the smaller. So I tried to set $AE$ to $x$ and $EC$ as $\dfrac{4}{3}$$x$ and then using Pythagoras to find the lenght of $AB$. Because if i have $AB$ I know that $AF$ = $\dfrac{3}{4}$$FB$ which I then could use to find the height in the $\triangle$$EFB$ but I just cant seem to get there. I always have to many unknowns in my equation, what am I missing?
Best regards
$$\frac{EF}9=\frac{FB}{AB}\tag{1}$$
$$\frac{EF}{12}=\frac{AF}{AB}=\frac{AB-FB}{AB}=1-\frac{FB}{AB}\tag{2}$$
By adding (1) and (2) you get:
$$\frac{EF}9+\frac{EF}{12}=1$$
You can easily solve this equation for $EF$.