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?
Best Answer
See that $D$ is the $A-$excenter of $\triangle AEF$ wherefrom $\widehat{EDF}=90^\circ-\dfrac{\widehat{BAC}}2$, i.e. $\widehat{CDF}=\dfrac{\widehat{BAC}}2$.