A generalization of Bottema's theorem 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.
 A: 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
$$H={1\over2}(F+G)={1\over2}(A+B)+{\lambda i\over2}(B-A)\ .$$
A: 
Basically here we mimic the proof from cut-the-knot, but replacing congruence with similarity:
Let
$$\frac {AH}{AG} = \frac {BI}{BD} = a$$
Let's drop perpendiculars $HL$, $CX$, $JK$, and $IM$ onto $AB$. (I forgot to label $X$)
$JK$ is the midline of trapezoid $HLMI$ so that
$$JK = \frac {HL + IM}2$$
Further, since $\angle HAC$ is right, $\angle HAL$ and $\angle CAX$ are complementary which makes right triangles $\triangle HAL$ and $\triangle ACX$ similar, implying
$$HL=aAX$$
Similarly,
$$IM=aBX$$
Taking all three identities into account shows that
$$JK = \frac {HL + IM}2 = \frac a2 (AX+BX) = \frac a2 AB = aAK$$
independent of $C$. No trigs but tricks.

EDIT: I see how I can prove that $AK=KB$.
By the previous similar triangles ($\triangle HAL \sim \triangle ACX$ and $\triangle IBM \sim \triangle BCX$), we have:
$$AL = k CX = BM$$
By midline theorem, $LK = KM$.
Therefore $AK = LK-AL=KM-BM=KB$.
This shows that (finally!) $J$ is fixed, since it is at a fixed distance "above" the midpoint of $AB$.
A: A little angle-chasing shows that the target point (here, $K$) is the midpoint of a side of a particular, symmetrically-situated parallelogram, which in turn shows that, for a given ratio $\lambda$, the point's position relative to side $\overline{AB}$ is independent of the position of $C$.

Note: $c$ is half of $|AB|$ in the figure.

FYI: If the right angles are formed "the other way" at $A$ and $B$, then the corresponding midpoint is the reflection of $K$ across $\overline{AB}$. Proof is left as an exercise to the reader.
A: Here is a proof via vectors. This avoids the issue of the location of $J$ in my earlier proof.
Use the original diagram and let $O$ be the midpoint of $AB$.
Let $\overrightarrow {OB} = a \hat i$. Then $\overrightarrow {OA} = -a \hat i$.
Let $\overrightarrow {OC} = b \hat i + c \hat j$.
Hence $\overrightarrow {AC} = (a+b) \hat i + c \hat j$ and $\overrightarrow {BC} = (-a+b) \hat i + c \hat j$.
We can easily show that $\overrightarrow {AE} = -c \hat i+(a+b) \hat j $ and $\overrightarrow {BD} = c \hat i + (a-b) \hat j$.
Letting $\dfrac {AF}{AE} = \dfrac {BG}{BD} = k$ we have $\overrightarrow {AF} = -kc \hat i+k(a+b) \hat j $ and $\overrightarrow {BG} = kc \hat i + k(a-b) \hat j$.
Finally:
$$\begin{align}\overrightarrow{OH}&=\frac12(\overrightarrow{OF}+\overrightarrow{OG})\\&=\frac12(\overrightarrow{OA}+\overrightarrow{AF}+\overrightarrow{OB}+\overrightarrow{BG})
\\&=\frac12(-a\hat i-kc \hat i+k(a+b) \hat j+a\hat i+kc \hat i + k(a-b) \hat j)
\\&=\frac k2((a+b)+(a-b))\hat j
\\&=ka\hat j
\end{align}$$
This shows that $OH \perp AB$ and $|OH|$ only depend on $a$ and $k$, that is, the length of $AB$ and the ratio $k$, implying the position of $H$ is indeed fixed.
