Let $ABCD$ be a convex quadrilateral with $AB>AD$ and $\angle B=\angle D=90^{\circ}$. Let $P$ be a point in the side $AB$ such that $AP=AD$. The lines $PD$ and $BC$ cut in the point $Q$. The perpendicular line to $AC$ passing by $Q$ cuts $AB$ in the point $R$. Let $S$ be the foot of perpendicular of $D$ to the line $AC$. Prove that $\angle PSQ=\angle RCP$.
Problem
Source: Rioplatense L3 2023 #2
Tags: geometry
v4913
08.12.2023 05:36
First, let $E$ be the reflection of $D$ over $AC$; then $A$ is the circumcenter of $(EPD)$. Claim 1: Let $K = DP \cap (ABCD)$; then $K$ is the circumcenter of $(PBQE)$. Proof: $\angle{EAP} = 2\angle{EDP} \implies K$ is the midpoint of $\widehat{EB}$, so $KE = KB$, and $\angle{EKP} = 2\angle{EBP}$ so $K$ is the circumcenter of $(PQ)$. Also, $\triangle{EPQ} \sim \triangle{EAC} \implies \triangle{EAP} \sim \triangle{ECQ} \implies CE = CQ = CD$. Let $Q'$ be the $Q$-antipode wrt $(QED)$; then $E, P, Q'$ are collinear, and $\triangle{PED} \sim \triangle{PQQ'}$. Now if $Q''$ is such that $\triangle{QED} \sim \triangle{Q''QQ'}$, then $Q''$ is on $PQ'$ such that $QQ'' \perp QD$, and $\angle{PSQ} = \angle{PCQ''}$, so it suffices to show that $C, R, Q''$ are collinear. Claim 2: If $X = CR \cap (ABCD)$, then $Q''QXE$ is cyclic. Proof: $\angle{EQ''Q} = \angle{EQ'D} = \angle{ECA} = \angle{EXA}$. Thus, $\angle{CXQ} = \angle{Q''XQ} = 90^{\circ} \implies C, R, X, Q''$ are collinear, as desired. $\square$
Scilyse
02.01.2025 10:06
this problem ruined my geo exam perfect score