Prove that SD=IE

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Prove that SD=IE

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Source: Kyiv mathematical festival 2021

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rogue

#1 h May 8, 2021, 10:44 PM • 3 Y

Y by Mango247, Mango247, Mango247

Let $\omega$ be the circumcircle of a triangle $ABC$ ( ${AB\ne AC}$ ), $I$ be the incenter, $P$ be the point on $\omega$ for which $\angle API=90^\circ,$ $S$ be the intersection point of lines $AP$ and $BC,$ $W$ be the intersection point of line $AI$ and $\omega.$ Line which passes through point $W$ orthogonally to $AW$ meets $AP$ and $BC$ at points $D$ and $E$ respectively. Prove that $SD=IE.$ (Ye. Azarov)

This post has been edited 1 time. Last edited by rogue, May 8, 2021, 10:45 PM

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zuss77

#2 h May 9, 2021, 9:26 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $T$ be foot of $I$ on $BC$ . We see that inversion $(W;WI)$ maps $P\mapsto T$ . And we have two results:
a) $(SPIT)$ is fixed, $WI$ is tangent to it, $SI\parallel DE$ .
b) $T\in PW$ . So it's a Simson line of $I$ and hence $I\in (DSE)$ .
With all this $DSIE$ - isoscales trapezoid.

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#3 h May 12, 2021, 9:23 PM

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Let $A'$ , $K$ be the antipode of $A$ on $(ABC)$ and the foot of the perpendicular from $I$ to $\overline{BC}$ , respectively. It's well-known that $P \equiv \overline{WK} \cap \overline{A'I}$ . By the radical center theorem for $(ABC)$ , $(AIP)$ and the incircle of $\triangle ABC$ we get that $\overline{SI}$ is tangent to $(BIC)$ , and thus is perpendicular to $\overline{AW}$ and parallel to $\overline{DW}$ . So, $\frac{SD}{AS}=\frac{WI}{AI}$ , so it suffices to prove that $\frac{IE}{IW}=\frac{AS}{AI}$ . For this notice the cyclic quadrilaterals $IKWE$ and $IKPS$ , and so $\angle WIE=\angle WKE=\angle PKS=\angle PIS=\angle SAI$ , so $\triangle SAI \sim \triangle EIW$ , and the conclusion follows. $\blacksquare$

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