The altitude of one triangle is the median of the other.


Let the squares have sides $a$ and $b$ and consider the altitude to triangle ABC in the diagram. Then by the sine rule

${\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}.}$

The altitude to triangle ABC cuts the angle at the common vertex in the other triangle into two angles equal to angles $A$ and $B$ and the opposite side into lengths $p$ and $q$ where

${\displaystyle p=\frac{b\sin A}{\sin z}=\frac{a\sin B}{\sin z}=\frac{a\sin B}{\sin (\pi -z)}=q}$

showing that this is the median of the triangle.