Let triangle $ABC$ be an acute triangle with $AB\neq AC$. The bisector of $BC$ intersects the lines $AB$ and $AC$ at points $F$ and $E$, respectively. The circumcircle of triangle $AEF$ has center $P$ and intersects the circumcircle of triangle $ABC$ at point $D$ with $D$ different to $A$. Prove that the line $PD$ is tangent to the circumcircle of triangle $ABC$.
Problem
Source: Lusophon Mathematical Olympiad 2021 Problem 3
Tags: geometry, circumcircle
teddy8732
20.12.2021 05:20
WLOG $AB>AC$ Since $ \angle AEF=90-C$ $ \rightarrow $ $ \angle PAF = C = \angle ACB $ So $PA$ is tangent to circumcircle of $ABC$ So $PD$ is also tangent to circumcircle of $ABC$.
Rafinha
20.12.2021 05:41
Consider an inversion around the circumcircle of $\triangle ABC$. Let $F '$ be the inverse of $E$, note that, with $O$ being the center of $(ABC)$, we have that, as $E$ is in the perpendicular bisector of $BC$, then $\angle OCE=\angle OAE=\angle OBE \Rightarrow ABOE$ is cyclic, therefore $A,B,F'$ are collinear $\Rightarrow F'=F$. So we have that $(ABC)$ and $(AEF)$ are orthogonal and the result follows.
Jalil_Huseynov
20.12.2021 10:57
Let $O$ be center of $(ABC) \implies \angle APE=2\angle AFE=2(90-\angle ABC) \implies \angle PAC=\angle ABC \implies PA$ tangents to $(O)$. Since $AD$ is polar of $P$ $\text{wrt}$ $(O)$, we get $PD$ tangents to $(O)$,too.
Mahdi_Mashayekhi
20.12.2021 11:19
∠AFD = ∠CBD + 90 - ∠B ---> ∠PDA = ∠B - ∠CBD = ∠DBA ---> PD is tangent to circumcircle of triangle ABC.