Let $ABC$ an isosceles triangle with $CA=CB$ and $\Gamma$ it´s circumcircle. The perpendicular to $CB$ through $B$ intersect $\Gamma$ in points $B$ and $E$. The parallel to $BC$ through $A$ intersect $\Gamma$ in points $A$ and $D$. Let $F$ the intersection of $ED$ and $BC, I$ the intersection of $BD$ and $EC, \Omega$ the cricumcircle of the triangle $ADI$ and $\Phi$ the circumcircle of $BEF$.If $O$ and $P$ are the centers of $\Gamma$ and $\Phi$, respectively, prove that $OP$ is tangent to $\Omega$
Problem
Source: Mathematics Regional Olympiad of Mexico Southeast 2018 P5
Tags: geometry, circumcircle
UI_MathZ_25
24.11.2022 08:42
It's clear that $C, O$ and $E$ are collinear. Claim: $O$ lies on $\Omega$. Proof: $\angle AOI = 2 \angle ACE = \angle ACB = 180^{\circ} - \angle ADB = 180^{\circ} - \angle ADI$ $\square$ Since $PO \perp BE \Rightarrow PO \parallel BC \parallel AD$ and as $O$ lies on the perpendicular bisector of $AD$, therefore $OP$ is tangent to $\Omega$ at $O$, as desired.
Attachments:
Proproblemsolver
24.11.2022 09:10
UI_MathZ_25 wrote: It's clear that $C, O$ and $E$ are collinear. Claim: $O$ lies on $\Omega$. Proof: $\angle AOI = 2 \angle CAE = \angle ACB = 180^{\circ} - \angle ADB = 180^{\circ} - \angle ADI$ $\square$ Since $PO \perp BE \Rightarrow PO \parallel BC \parallel AD$ and as $O$ lies on the perpendicular bisector of $AD$, therefore $OP$ is tangent to $\Omega$ at $O$, as desired. How $ \angle AOI$ = 2$ \angle CAE$ ?
UI_MathZ_25
26.11.2022 01:16
Proproblemsolver wrote: UI_MathZ_25 wrote: It's clear that $C, O$ and $E$ are collinear. Claim: $O$ lies on $\Omega$. Proof: $\angle AOI = 2 \angle CAE = \angle ACB = 180^{\circ} - \angle ADB = 180^{\circ} - \angle ADI$ $\square$ Since $PO \perp BE \Rightarrow PO \parallel BC \parallel AD$ and as $O$ lies on the perpendicular bisector of $AD$, therefore $OP$ is tangent to $\Omega$ at $O$, as desired. How $ \angle AOI$ = 2$ \angle CAE$ ? Done