I was attempting the 2001 BMO 2 and was unable to solve question 3. The question was:
A triangle ABC has $\measuredangle ACB > \measuredangle ABC$. The internal bisector of $\measuredangle BAC$ meets BC at D. The point E on AB is such that $\measuredangle EDB = 90◦$. The point F on AC is such that $\measuredangle BED = \measuredangle DEF$. Show that $\measuredangle BAD = \measuredangle FDC$.
My progress:
I first started out labelling $\measuredangle BAD = \measuredangle DAC = \alpha$ and $\measuredangle ABC = \beta$. I then did some angle chasing but found nothing interesting like a cyclic quadrilateral or parallel lines. However, I did find that FE was perpendicular to AB. I then tried to create some cyclic quadrilaterals. I chose to look at the $\triangle EFA$ first, so I drew the perpendicular from F to AD and set its foot as X. This way we have AEXF as a cyclic quadrilateral with diameter AF. Using angles in the same segment we see $\measuredangle XFE = \measuredangle XEF = \alpha$ so XE = XF. Then I drew the line XY from X to AB such that $\measuredangle XYA = \measuredangle XAY = \alpha$ which then gives XA = XY. From there, with some angle chasing you can also find that $\triangle XEY$ is a right angled triangle and is congruent to $\triangle FXA$, also $\triangle XEF$ is similar to $\triangle XYA$.
From here I couldn't find anything useful though. How would someone continue along this path or have I missed an easier path before? I have learned a little about spiral similarity before, here, would $\triangle XAY$ and $\triangle XFE$ have spiral similarity around X of 90◦? Could that be used in any way?
Let's draw a line from $F$ which is parallel to $CB$, intersecting $AB$ at $G$. Since $ED$ is perpendicular to $FG$, and $\angle FED=\angle GED$, we can easily conclude that $FD=DG$. Now, let's assume $\angle ADF=y_1$, $\angle ADG=y_2$, and $\angle FDC=\angle GDB=x.$ By the law of sine in $\triangle FCD$, we have:
$$\frac{\sin (y_1+\frac{\angle A}{2})}{\sin x}=\frac{CD}{CF},$$ and in $\triangle GDB$,
$$\frac{\sin (y_2+\frac{\angle A}{2})}{\sin x}=\frac{DB}{GB}.$$ Since $FG$ and $BC$ are parallel and $AD$ is the angle bisector, we have: $\frac{CD}{CF}=\frac{DB}{GB}$.
Therefore, $\sin (y_1+\frac{\angle A}{2})=\sin(y_2+\frac{\angle A}{2}).$ So, $y_1+y_2+\frac{\angle A}{2}+\frac{\angle A}{2}=180^{\circ}.$ Hence, $AGDF$ is cyclic.
The rest is easy.