Let $AC$ and $BD$ intersect at $O$. Then, points $A,F,G,B,O$ are concyclic on the circle with diameter $AB$. Hence, $\angle OFE=\angle ABO=\angle ODE,$ which implies that $O \in (FDE)$. Similarly, $O \in (GEC)$. Therefore, $\angle HFO=\angle OED=\angle OGC,$ which implies that $H \in (FOG)$, hence $H \in (AB)$, which implies the desired result.

Here is a Cartesian coordinate approach to the problem. [asy][asy] size(9cm); pair A = dir(135); pair B = dir(225); pair C = dir(315); pair D = dir(45); pair E = 0.7*C + 0.3*D; pair F = foot(B,A,E); pair G = foot(A,B,E); pair H = extension(D,F,C,G); draw(A--B--C--D--A--cycle); draw(A--E--B); draw(D--H--C); draw(A--G); draw(B--F); draw(A--H--B,dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); [/asy][/asy] Let $A = (0,1)$, $B=(0,0)$, $C=(1,0)$ and $D=(1,1)$. Define $E=(1,e)$. First, we find the coordinate of $F$. The gradient of $AE$ is $e-1$, therefore the gradient of $BF$ is $\frac{1}{1-e}$. The equation of $AE$ is $y=(e-1)x+1$, and therefore if $F=\left(f,\frac{f}{1-e}\right)$, then we have \[\frac{f}{1-e}=(e-1)f+1\]\[f=(2e-e^2-1)f+1-e\]\[f=\frac{1-e}{e^2-2e+2}\] Now we find the coordinate of $G$. The gradient of $BE$ is $e$, thus the gradient of $AG$ is $-\frac{1}{e}$ and the equation of $AG$ is $y=-\frac{1}{e}x+1$. If $G=(g,eg)$, then \[eg=-\frac{g}{e}+1\]\[g=\frac{e}{e^2+1}\] We now find the equations of line $FD$ and $CG$. The gradient of $FD$ is \[\frac{1-\frac{f}{1-e}}{1-f}=\frac{1-\frac{1}{e^2-2e+2}}{1-\frac{1-e}{e^2-2e+2}}=\frac{e^2-2e+1}{e^2-e+1}\]Therefore, the equation of $FD$ is \[y=\frac{e^2-2e+1}{e^2-e+1}x+\frac{e}{e^2-e+1}.\] The gradient of $CG$ is \[-\frac{eg}{1-g}=-\frac{\frac{e^2}{e^2+1}}{\frac{1-e+e^2}{e^2+1}}=-\frac{e^2}{e^2-e+1}\]Therefore, the equation of $CG$ is \[y=-\frac{e^2}{e^2-e+1}x+\frac{e^2}{e^2-e+1}.\] Now, we intersect $CG$ and $FD$. \[\frac{e^2-2e+1}{e^2-e+1}x+\frac{e}{e^2-e+1}=-\frac{e^2}{e^2-e+1}x+\frac{e^2}{e^2-e+1}\]Multiplying throughout by $e^2-e+1$, we have \[x(2e^2-2e+1)=e^2-e\]\[x=\frac{e(e-1)}{2e^2-2e+1}\]\[y=\frac{e^2}{e^2-e+1}(1-x)=\frac{e^2}{e^2-e+1}\left(\frac{e^2-e+1}{2e^2-2e+1}\right)=\frac{e^2}{2e^2-2e+1}.\]The gradient of $AH$ is thus \[\frac{1-\frac{e^2}{2e^2-2e+1}}{\frac{e-e^2}{2e^2-2e+1}}=\frac{e^2-2e+1}{e(1-e)}=\frac{1-e}{e}.\]On the other hand, the gradient of $BH$ is \[\frac{e^2}{e^2-e}=\frac{e}{e-1}.\]Since the gradient of $AH$ is negative of the reciprocal of the gradient of $BH$, we have $\angle AHB=90^\circ$ and we are done.