Problem: Let $ABCD$ be a convex quadrilateral inscribed in a circle and satisfying
\[ DA < AB = BC < CD. \]
Points $E$ and $F$ are chosen on sides $CD$ and $AB$
such that $\ol{BE} \perp \ol{AC}$ and $\ol{EF} \parallel \ol{BC}$.
Prove that $FB=FD$.

Solution: We present three approaches.
We note that in the second two approaches,
the result remains valid even if $AB \neq BC$,
as long $E$ is replaced by the point on $\ol{AC}$
satisfying $EA = EC$.
So the result is actually somewhat more general.

\paragraph{First solution by inscribed angle theorem.}
Since $\ol{EF} \parallel \ol{BC}$ we may
set $\theta = \angle FEB = \angle CBE = \angle EBF$.
This already implies $FE = FB$,
so we will in fact prove that $F$ is the circumcenter of $\triangle BED$.
\begin{center}
\begin{asy}
pair A = dir(175);
pair B = dir(270);
pair C = B*B/A;
pair D = dir(135);
pair E = extension(B, origin, C, D);
pair F = circumcenter(B, D, E);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(D--F--B, lightred);
draw(F--E, lightred);
draw(B--D--C, blue);
draw(F--A--C, blue);
draw(E--B--C, blue);
draw(A--D, blue);
draw(CP(F, E), lightred);

markangle(1,7.0,E,B,F,deepgreen);
markangle(1,9.0,C,B,E,deepgreen);
markangle(1,9.0,F,E,B,deepgreen);

markangle(2,9.0,B,A,C,deepcyan);
markangle(2,9.0,B,D,C,deepcyan);
markangle(2,9.0,A,C,B,deepcyan);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(90));
dot("$F$", F, dir(250));

/* TSQ Source:

A = dir 175
B = dir 270
C = B*B/A
D = dir 135
E = extension B origin C D R90
F = circumcenter B D E R250
unitcircle 0.1 lightcyan / blue
D--F--B lightred
F--E lightred
B--D--C blue
F--A--C blue
E--B--C blue
A--D blue
CP F E lightred

!markangle(1,7.0,E,B,F,deepgreen);
!markangle(1,9.0,C,B,E,deepgreen);
!markangle(1,9.0,F,E,B,deepgreen);

!markangle(2,9.0,B,A,C,deepcyan);
!markangle(2,9.0,B,D,C,deepcyan);
!markangle(2,9.0,A,C,B,deepcyan);

*/
\end{asy}
\end{center}
Note that $\angle BDC = \angle BAC = 90\dg - \theta$.
However, $\angle BFE = 180\dg - 2 \theta$.
So by the inscribed angle theorem, $D$ lies on the circle
centered at $F$ with radius $FE = FB$, as desired.

\begin{remark*}
  Another approach to the given problem is to show that $B$
  is the $D$-excenter of $\triangle DAE$,
  and $F$ is the arc midpoint of $\widehat{DAE}$
  of the circumcircle of $\triangle DAE$.
  In my opinion, this approach is much clumsier.
\end{remark*}

\paragraph{Second general solution by angle chasing.}
By Reim's theorem, $AFED$ is cyclic.
\begin{center}
\begin{asy}
pair A = dir(175);
pair B = dir(255);
pair C = dir(5);
pair D = dir(135);
pair E = extension(B, origin, C, D);
pair F = extension(A, B, E, E+B-C);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(D--F, lightblue);
draw(A--E, lightblue);
draw(F--B--D--C, blue);
draw(F--A--C, blue);
draw(E--B--C, blue);
draw(A--D, blue);
draw(E--F, lightred);

draw(circumcircle(E, A, D), lightred);

markangle(2,13.0,C,A,E,deepgreen);
markangle(2,13.0,E,C,A,deepgreen);
markangle(2,13.0,D,B,F,deepgreen);
markangle(2,13.0,F,D,B,deepgreen);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(70));
dot("$F$", F, dir(F));

/* TSQ Source:

A = dir 175
B = dir 255
C = dir 5
D = dir 135
E = extension B origin C D R70
F = extension A B E E+B-C
unitcircle 0.1 lightcyan / blue
D--F lightblue
A--E lightblue
F--B--D--C blue
F--A--C blue
E--B--C blue
A--D blue
E--F lightred

circumcircle E A D lightred

!markangle(2,13.0,C,A,E,deepgreen);
!markangle(2,13.0,E,C,A,deepgreen);
!markangle(2,13.0,D,B,F,deepgreen);
!markangle(2,13.0,F,D,B,deepgreen);

*/
\end{asy}
\end{center}
Hence
\begin{align*}
  \dang FDB &= \dang FDC - \dang BDC = \dang FAE - \dang FAC \\
  &= \dang CAE = \dang ECA = \dang DCA = \dang DBA = \dang DBF
\end{align*}
as desired.


\paragraph{Third general solution by Pascal.}
Extend rays $AE$ and $DF$ to meet the circumcircle again at $G$ and $H$.
By Pascal's theorem on $HDCBAG$,
it follows that $E$, $F$, and $GH \cap BC$ are collinear,
which means that $\ol{EF} \parallel \ol{GH} \parallel \ol{BC}$.
\begin{center}
\begin{asy}
pair A = dir(175);
pair B = dir(255);
pair C = dir(5);
pair D = dir(135);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair H = D*B/A;
pair G = A*C/D;
pair E = extension(A, G, D, C);
pair F = extension(A, B, D, H);

draw(A--B--C--D--cycle, blue);
draw(A--G, deepgreen);
draw(D--H, deepgreen);
draw(E--F, red);
draw(G--H, red);

draw(arc(origin,C,G), orange+1.5);
draw(arc(origin,D,A), orange+1.5);
draw(arc(origin,H,B), orange+1.5);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$H$", H, dir(H));
dot("$G$", G, dir(G));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));

/* TSQ Source:

A = dir 175
B = dir 255
C = dir 5
D = dir 135
unitcircle 0.1 lightcyan / blue
H = D*B/A
G = A*C/D
E = extension A G D C
F = extension A B D H

A--B--C--D--cycle blue
A--G deepgreen
D--H deepgreen
E--F red
G--H red

!draw(arc(origin,C,G), orange+1.5);
!draw(arc(origin,D,A), orange+1.5);
!draw(arc(origin,H,B), orange+1.5);

*/
\end{asy}
\end{center}
Since $EA=EC$, it follows $DAGC$ in isosceles trapezoid.
But also $GHBC$ is an isosceles trapezoid.
Thus $\mathrm{m}\widehat{DA} = \mathrm{m}\widehat{GC}
= \mathrm{m}\widehat{BH}$,
so $DAHB$ is an isosceles trapezoid.
Thus $FD = FB$.

\begin{remark*}
  Addicts of projective geometry can use Pascal
  on $DBCAHG$ to finish rather than noting the equal arcs.
\end{remark*}