Let $PQRS$ be a quadrilateral with $| P Q | = | QR | = | RS |$, $\angle Q= 110^o$ and $\angle R = 130^o$ . Determine $\angle P$ and $\angle S$ .
Problem
Source: 2014 VWO Flanders MO p3
Tags: geometry, equal segments, angles, Angle Chasing
rafaello
07.08.2020 16:38
Reflecting $R$ from line $SP$, thus we know that $SR'=PQ$ and $P,Q,S,R'$ are on the same circle, thus $PQSR'$ is an isosceles trapezoid, which means that $PQS=QSR'=85^{\circ}$, which means that $\angle RSP=\frac{\angle RSR'}{2}=\frac{25^{\circ}+85^{\circ}}{2}=55^{\circ}$, thus $\angle SPQ= 360^{\circ}-\angle PQR-\angle QRS - \angle RSP=360^{\circ}-110^{\circ}-130^{\circ} - 55^{\circ}=65^{\circ}.$
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