BP=PQ wanted, CD=CE, right isosceles (2021 Novosibirsk Oral Geo Oly 8.5)

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BP=PQ wanted, CD=CE, right isosceles (2021 Novosibirsk Oral Geo Oly 8.5)

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#1 h May 24, 2021, 5:07 PM • 3 Y

Y by centslordm, Mango247, Mango247

On the legs $AC$ and $BC$ of an isosceles right-angled triangle with a right angle $C$ , points $D$ and $E$ are taken, respectively, so that $CD = CE$ . Perpendiculars on line $AE$ from points $C$ and $D$ intersect segment $AB$ at points $P$ and $Q$ , respectively. Prove that $BP = PQ$ .

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#2 h May 24, 2021, 9:57 PM

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We unfortunately use coordinates :mad:

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. Scale the figure such that $C=(0,0)$ , $A=(0,1)$ , and $B=(1,0)$ . Then $E=(a,0)$ and $D=(0,a)$ . We then have $Q=\left(\frac{1-a}{a+1},\frac{2a}{a+1}\right)$ $P=\left(\frac{1}{a+1},\frac{a}{a+1}\right).$ We have $PQ-BP$ $= \sqrt{\left(\frac{a}{a+1}-\frac{2a}{a+1}\right)^2+\left(\frac{1}{1+a}-\frac{1-a}{a+1}\right)^2}-\sqrt{\left(0-\frac{a}{a+1}\right)^2+\left(1-\frac{1}{1+a}\right)^2}$ $= \frac{a\sqrt{2}}{a+1}-\frac{a\sqrt{2}}{a+1} = 0$ $\implies BP=PQ$ $\blacksquare$

This post has been edited 1 time. Last edited by Honestly, May 30, 2021, 6:18 PM

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