prove PQS is isosceles, PQ ^ 2 = QR \cdot ST

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prove PQS is isosceles, PQ ^ 2 = QR \cdot ST

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Source: Kazakhstan NMO 2007 grade X P2

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#1 h Sep 28, 2018, 7:25 AM • 1 Y

Y by Adventure10

The acute triangle $ABC$ is inscribed in a circle with the center $O$ . The point $P$ is chosen on the smaller of the two arcs $AB$ . The line passing through $P$ perpendicular on $BO$ intersects the sides $AB$ and $BC$ at the points $S$ and $T$ , respectively. The line passing through $P$ perpendicular on $AO$ intersects the sides $AB$ and $AC$ at points $Q$ and $R$ , respectively. Prove that:
a) the triangle $PQS$ is isosceles;
b) $PQ ^ 2 = QR \cdot ST.$

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#2 h Sep 28, 2018, 8:33 AM • 2 Y

Y by Adventure10, Mango247

Let the intersection of $PR,AO$ be $M$ . Define $N$ on $BO$ similarly.

Then, since $\angle OAB=\angle OBA, \angle AQM=\angle BSN$ .

Now, $\angle PSQ=\angle BSN=\angle AQM=\angle PQS$ . Thus $PQS$ is isosceles.

Since $\angle AOB=2C, \angle AQM=\angle BSN=90-C$ . So $\angle BST=\angle AQR=C$ . This implies $\Delta AQR\sim \Delta TSB$ . Thus $\frac{QR}{BS}=\frac{AQ}{ST}$ . Thus $QR\cdot ST=AQ\cdot BS$ .

Let $\angle BPS=x$ , so $\angle APQ=C-x$ . Since $\angle PSB=\angle PQA=180-C$ , $\angle PAQ=x,\angle PBS=C-x$ . So $\Delta BSP\sim\Delta PQA$ . So $\frac{BS}{PS}=\frac{PQ}{AQ}$ .

Since $PQ=PS$ , therefore $PQ^2=AQ\cdot BS=QR\cdot ST$ .

$\tiny\boxed Z$

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