cyclic wanted, AP: PB = AT: TC = 1: 2 in a right ABC

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cyclic wanted, AP: PB = AT: TC = 1: 2 in a right ABC

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Source: All-Siberian Open School Olympiad 2018-19 11.4

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#1 h Jul 6, 2020, 7:02 AM

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In a right triangle $ABC$ , point $M$ is the midpoint of the hypotenuse $BC$ , and the points $P$ and $T$ divide the legs AB and AC in ratios $AP: PB = AT: TC = 1: 2$ . We denote by $K$ the intersection point of the segments $BT$ and $PM$ , and by $E$ the intersection point of the segments of $CP$ and $MT$ , and by $O$ the intersection point of the segments of $CP$ and $BT$ . Prove that the quadrilateral $OKME$ is cyclic.

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#2 h Jul 6, 2020, 3:02 PM

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First notice by Menelaus on $\triangle ABC$ we have that $BT, CP, AM$ concur at $O$ , now let $M'=AM \cap PT$ and this is the midpoint of $PT$ , so again Menelaus on $\triangle MPT$ we have $KE \| PT$ . Let $N$ be the midpoint of $PB$ , see that $\triangle BNM \cong \triangle APM$ , then $\angle BMN=\angle BCP=\angle KEO=\angle AMP=\angle KMO \blacksquare$

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