Given a square $ABCD$, points $E$ and $F$ are taken inside the segments $BC$ and $CD$ so that $\angle EAF = 45^o$. The lines $AE$ and $AF$ intersect the circle circumscribed to the square at points $G$ and $H$ respectively. Prove that lines $EF$ and $GH$ are parallel.
WLOG $s=1$ ,s is the side
$FC=a$
$EC=b$
then by intersecting chord theorem
$EG = \frac{(1-b)*b}{1+(1-b)^2}$
$FH = \frac{(1-a)*a}{1+(1-a)^2}$
by thales
$\frac{AF}{FH} = \frac{AE}{EG}$
$\frac{(1+(1-a)^2)}{(1-a)*(a)} = \frac{(1+(1-b)^2)}{(1-b)*(b)} $
to prove this we need some bash since $EAF=45$
cosine rule
$a^2+b^2 = (1+(1-a)^2)+(1+(1-b)^2)-2*\sqrt{1+(1-a)^2}*\sqrt{1+(1-b)^2}*\frac{\sqrt{2}}{2}$
which simplify to
$2(a+b)(2-ab)+(ab-2)(ab+2)=0$
$(2-ab)(2a+2b-ab-2)=0$
(2-ab)>0 then (2a+2b-ab-2) is the roots
Continuing this process
$\frac{(1+(1-a)^2)}{(1-a)*(a)} = \frac{(1+(1-b)^2)}{(1-b)*(b)} $
$2(b-a)=2(b-a)(b+a)-ab(b-a)$
$(b-a)(2-2b-2a+ab)=0$