equal angles wanted, 2 circles, angle bisector related

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Source: 2019 Saudi Arabia IMO Training Test 5.2

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#1 h Sep 4, 2020, 7:41 AM

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Let the bisector of the outside angle of $A$ of triangle $ABC$ and the circumcircle of triangle $ABC$ meet at point $P$ . The circle passing through points $A$ and $P$ intersects segments $BP$ and $CP$ at points $E$ and $F$ respectively. Let $AD$ is the angle bisector of triangle $ABC$ . Prove that $\angle PED = \angle PFD$ .

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This post has been edited 8 times. Last edited by parmenides51, Jan 5, 2022, 3:35 PM
Reason: added image, and corrected typo

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#2 h Jan 5, 2022, 3:31 PM

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posted for the image link, as I was trying to find the typo in the original wording
instead of $AD$ angle bisector of $ABC$ , it was written on the official pdf that $AD$ was bisector of $AD$

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#3 h Jan 5, 2022, 3:50 PM

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It is easy to see that $A$ is the Miquel point of $B,C,F,E,P,EF\cap BC$ so it is the centre of the spiral similarity which sends $EB\to FC$ and also the centre of the similarity which sends $EF\to BC$ .From that we get $\Delta AEB\sim\Delta AFC$ and $\Delta AEF\sim\Delta ABC$ .Thus $\frac{BE}{FC}=\frac{AE}{AF}=\frac{AB}{AC}=\frac{BD}{DC}$ , but $\angle EBD=\angle FCD$ , therefore $\Delta BED\sim\Delta CFD\Rightarrow \angle PED=\angle PFD$ .

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