A point $P$ was chosen on the smaller arc $BC$ of the circumcircle of the acute-angled triangle $ABC$. Points $R$ and $S$ on the sides$ AB$ and $AC$ are respectively selected so that $CPRS$ is a parallelogram. Point $T$ on the arc $AC$ of the circumscribed circle of $\vartriangle ABC$ such that $BT \parallel CP$. Prove that $\angle TSC = \angle BAC$. (Anton Trygub)
Problem
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Tags: geometry, parallelogram, circumcircle, equal angles
collinator
22.09.2020 02:11
[asy][asy] size(200); import olympiad; pair A,B,C,P,R,S,T,X; A=dir(82); B=dir(-10); C=dir(-157); P=dir(-93); R=extension((A-C)+P,P,A,B); S=R+C-P; pair [] a=intersectionpoints(circumcircle(A,B,C),B--(B+10(C-P))); T=a[0]; pair[] b=intersectionpoints(circumcircle(A,B,C),T--(T+10(S-T))); draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(C--P--R--S); draw(B--T); draw(T--S); draw(T--b[0]); draw(P--b[0]); draw(T--C); draw(P--B); draw(A--b[0]); label("$A$",A,N); label("$B$",B,dir(-10)); label("$C$",C,dir(-157)); label("$P$",P,dir(-93)); label("$R$",R,E); label("$S$",S,N); label("$T$",T,NW); label("$X$",b[0],NE); [/asy][/asy] If we let $X$ be the intersection of line $PR$ with $(ABC)$ other than $P$, then since $\overline{PX}\parallel\overline{AC}$ and $PXAC$ cyclic, $PXAC$ is an isosceles trapezoid with $AX=PC$. Since $PCRS$ is a parallelogram, $SR=PC$ and so we can get that $SR=PC=AX$. Since $\overline{RX}\parallel \overline{SA}$ we see that $AXRS$ is an isosceles trapezoid, and thus cyclic, so $\measuredangle CAB=\measuredangle SAR=\measuredangle SXR$. Since $\overline{TB}\parallel\overline{CP}$ and $TBPC$ cyclic, $TBCP$ is an isosceles trapezoid with $\measuredangle CTB=\measuredangle TBP$ and since $TPBX$ cyclic, $\measuredangle TXR=\measuredangle TXP=\measuredangle TBP=\measuredangle CAB=\measuredangle SXR$ so $T,S,X$ collinear, and $\angle TSC=\angle ASX$ however since $AXRS$ is an isosceles trapezoid, $\angle TSC=\angle ASX=\angle RAS=\angle BAC$. In this solution $\measuredangle MNO$ represents directed angle $\angle MNO$.