TITLE: A generalization of Bottema's theorem
QUESTION [5 upvotes]: Can you provide a proof for the following claim:

In any triangle $ABC$ construct isosceles right triangles on sides $AC$ and $BC$, with right angles at the points $A$ and $B$. Let points $F$ and $G$ divide catheti $AE$ and $BD$ respectively in the same arbitrary ratio . The midpoint $H$ of the line segment that connects points $F$ and $G$ is independent of the location of $C$ .


GeoGebra applet that demonstrates this claim can be found here. I tried to mimic a proof of Bottema's theorem given on this page but without success.

REPLY [6 votes]: Consider $A$, $B$, $C$ as complex numbers, and choose a $\lambda\in{\mathbb R}$. Then
$$F=A+\lambda(E-A)=A+\lambda\,i(C-A),\qquad G=B+\lambda(D-B)=B+\lambda(-i) (C-B)\ .$$
It follows that
$$H={1\over2}(F+G)={1\over2}(A+B)+{\lambda i\over2}(B-A)\ .$$