equal segments starting with a rectangle and a circle

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Source: May Olympiad (Olimpiada de Mayo) 2010 L2

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#1 h Sep 11, 2018, 7:18 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a rectangle and the circle of center $D$ and radius $DA$ , which cuts the extension of the side $AD$ at point $P$ . Line $PC$ cuts the circle at point $Q$ and the extension of the side $AB$ at point $R$ . Show that $QB = BR$ .

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#2 h Sep 12, 2018, 12:07 AM • 2 Y

Y by Adventure10, Mango247

Note that $\triangle DCP$ , $\triangle BRC$ and $\triangle ABD$ are congruent, because them have two angles and a side in common.
Let $\theta =\angle DPC =\angle BCR =\angle ADB =\angle DQP$ . Then $\angle ABD =\angle BRC =\angle DCP =180-90-\theta=90-\theta$ .
As $DQ=AD=BC$ , $BD$ and $CQ$ are parallel and $\angle CBD=\angle QDB$ , $BDQC$ is an isosceles trapezium. Then $DC=BQ$ , so $\triangle BQD$ and $\triangle BDC$ are congruent. Then $\angle DQB=90$ . So $\angle BQC=180-90-\theta=90-\theta$ .
As $\angle BRQ=90-\theta =\angle BQR$ , $\boxed{QB=BR}$ .

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#3 h Sep 12, 2018, 4:21 AM • 2 Y

Y by Adventure10, Mango247

$AQ\bot PC\implies ABCQD$ cyclic, i.e. $\angle BQD=90^\circ$ , hence $BQ$ is tangent to the circle, thus $BQ=AB$ ; with $BR=AB$ we are done.

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#4 h Jun 10, 2024, 11:10 AM

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cyclic quads

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