We are given a convex quadrilateral $ABCD$ whose angles are not right. Assume there are points $P, Q, R, S$ on its sides $AB, BC, CD, DA$, respectively, such that $PS \parallel BD$, $SQ \perp BC$, $PR \perp CD$. Furthermore, assume that the lines $PR, SQ$, and $AC$ are concurrent. Prove thatthe points $P, Q, R, S$ are concyclic.
Problem
Source: MEMO 2023 T5
Tags: geometry
MarkBcc168
25.08.2023 12:10
Let $T=PR\cap SQ\cap AC$, and let $H$ be the orthocenter of $\triangle BCD$. Notice that $\triangle TPS$ and $\triangle HBD$ are homothetic, so $BP, DS, HT$ concur at $A$, which means that $H\in AC$, so $AC\perp BD$. Now, if $X$ is the foot from $T$ to $PR$, the result follows from $TP\cdot TR = TX\cdot TC = TQ\cdot TS$.
chrono_42
30.01.2024 00:34
Let $T = PR \cap SQ \cap AC$. Desargues' theorem, applied to the triangles $CDB$ and $TSP$, yields that $E = BC \cap PR$, $F = CD \cap SQ$ and $\infty_{BD} = BD \cap PS$ are collinear, so $EF \parallel BD \parallel PS$. Since $\angle FQE = \angle FRE = 90^\circ$, the quadrilateral $FERQ$ is cyclic. By Reim's theorem for the trapezoid $FEPS$ (or, if you will, $\angle RQS = 180^\circ - \angle FQR = \angle REF = \angle RPS$) we conclude that $PQRS$ is cyclic.
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