Two circles $O_1$ and $O_2$ intersect at $A,B$. Bisector of outer angle $\angle O_1AO_2$ intersects $O_1$ at $C$, $O_2$ at $D$. $P$ is a point on $\odot(BCD)$, $CP\cap O_1=E,DP\cap O_2=F$. Prove that $PE=PF$.
Problem
Source: 2019 China North MO, Problem 5
Tags: geometry
Pathological
22.02.2020 05:48
Observe that $\angle BPF = \angle BCD = \angle BEA = \angle BEF$ and so $BEPF$ is cyclic. This means that $B$ is the center of the spiral similarity sending $EC$ to $FD$. However, as $A = (\triangle ECB) \cap (\triangle FDB)$, we have that $A = CD \cap EF$ (well-known), and in particular $A \in EF$. Now, notice that $\angle PEF = \angle CBA = 90 - \angle O_1AC = 90 - \angle O_2AD = \angle DBA = \angle PFE$, and so we're done. $\square$
puntre
22.05.2021 16:17
Pathological wrote: $A = CD \cap EF$ (well-known) Does anyone have an explanation of this one?
llplp
22.05.2021 19:56
puntre wrote: Does anyone have an explanation of this one? $B$ is the centre of the spiral similarity sending $EC$ to $FD$ and so there exists another spiral similarity at $B$ sending $EF$ to $CD$. Hence $EF \cap CD$ is the intersection of $(BEC)$ and $(ADF)$, which is $A$.