Let $ABC$ be a triangle with incenter $I$. The line $BI$ meets $AC$ at $D$. Let $P$ be a point on $CI$ such that $DI=DP$ ($P\ne I$), $E$ the second intersection point of segment $BC$ with the circumcircle of $ABD$ and $Q$ the second intersection point of line $EP$ with the circumcircle of $AEC$. Prove that $\angle PDQ=90^\circ$. Proposed by Ariel García
Problem
Source: Mexico National Olympiad 2020 P2
Tags: geometry, incenter, circumcircle
Bassiskicking
12.11.2020 02:11
Step 1: $DP$ is tangent to $\odot ABD$ Proof: Use The Fact That $\angle BIC=90+\frac{1}{2} \cdot \angle BAC$ Step 2: $D$ lies in the perpendicular bisector of $AE$ Proof: $\angle ABD = \angle DBE$ Step 3: $CPDQ$ is cyclic: Proof: $ \angle PQC = \angle EQC =\angle EAC = \angle EAD =\angle DBE \stackrel{DP tangent}{=} \angle PDC $ Step 4: $Q$ lie in the perpendicular bisector of $AE$ Proof: $\angle AQE =\angle ACE = 2\angle DQP = 2\angle DQE$ Conclusion: The reflection of $E$ respect $DQ$ is $A$ so the circuncenter of $\odot ADE$ lies in $DQ$. So $\angle PDQ =90$.
Jjesus
14.11.2020 16:38
$\angle ABI=\alpha$ and $\angle IAE=\beta \Rightarrow \angle PDC=\alpha$ $\Rightarrow PCDQ$ is cyclic $\Rightarrow \angle PQD=90-2\alpha-\beta=\angle DQA$ $\Rightarrow DQ$ is perpendicular bisector $AE$ $\Rightarrow \angle DEP= \alpha+\beta \Rightarrow \angle PDQ=90^\circ$
cursed_tangent1434
25.01.2025 12:24
Solved with stillwater_25. Cool art problem! There is just one key observation to make after which the problem falls apart. Claim : Point $P$ is the incenter of $\triangle CDE$. Proof : Note that, \[\measuredangle PDB = \measuredangle PDI = 2\measuredangle PID = 2\measuredangle CIB = \measuredangle CAB = \measuredangle DAP\]which implies that $\overline{PD}$ is tangent to $(ADE)$. Now, this implies that \[\measuredangle PDE = \measuredangle DBE = \measuredangle DBC\]Also, \[\measuredangle CDP = \measuredangle ADB + \measuredangle IDP = \measuredangle CBA \]which implies that $P$ lies on the $\angle D-$internal angle bisector of $\triangle CDE$. It clear also lies on the $\angle C-$internal angle bisector which implies that $P$ is indeed the incenter, as desired. Claim : Point $Q$ is the $E-$excenter of $\triangle CDE$. Proof : It clearly lies on the $\angle E-$internal angle bisector due to the previous claim. Also, it lies on the external $\angle C-$bisector since, \[\measuredangle ECQ = \measuredangle CEQ + \measuredangle EQC = \measuredangle CEP + \measuredangle EAC = \measuredangle CEP + \measuredangle EBD = \measuredangle BAI + \measuredangle CBI\]Thus, by the Incenter-Excenter Lemma it follows that, points $P$ , $D$ , $Q$ and $C$ must be cyclic with $\angle PDQ = 90^\circ$.