PQ^2=QR= ST, isosceles, circumcircle related (2002 Brazilian L2 P5)

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PQ^2=QR= ST, isosceles, circumcircle related (2002 Brazilian L2 P5)

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#1 h Apr 24, 2020, 8:29 AM

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Let $ABC$ be a triangle inscribed in a circle of center $O$ and $P$ be a point on the arc $AB$ , that does not contain $C$ . The perpendicular drawn fom $P$ on line $BO$ intersects $AB$ at $S$ and $BC$ at $T$ . The perpendicular drawn from $P$ on line $AO$ intersects $AB$ at $Q$ and $AC$ at $R$ . Prove that:
a) $PQS$ is an isosceles triangle
b) $PQ^2=QR= ST$

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#2 h Apr 25, 2020, 7:24 PM

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$ST, RQ$ being perpendiculars onto $OB, OA$ are antiparalels in triangle, thus $\triangle AQR\sim\triangle TSB$ , giving $\frac{TS}{AQ}=\frac{SB}{RQ}$ , i.e. $TS\cdot RQ=AQ\cdot SB\ (\ 1\ )$ . From the antiparalels, we get $\widehat{PQS}=\widehat{PSQ}=\widehat{ACB}$ , thus a) has been solved.
Also note that $\triangle APQ\sim\triangle PBS$ ( if $\widehat{BAP}=\alpha$ , then $\widehat{ABP}=\widehat{ACB}-\alpha$ ) and from their similarity we get $\frac{PQ}{BS}=\frac{AQ}{PS}$ or $PQ\cdot PS=AQ\cdot BS\ (\ 2\ )$ , from $(1)$ and $(2)$ we get $PQ\cdot PS=QR\cdot ST$ .

Best regards,
sunken rock

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