right angle wanted, circumcircle of a right triangle related

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Source: 2015 Saudi Arabia JBMO TST 1.3

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#1 h Jul 21, 2020, 8:14 AM

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A right triangle $ABC$ with $\angle C=90^o$ is inscribed in a circle. Suppose that $K$ is the midpoint of the arc $BC$ that does not contain $A$ . Let $N$ be the midpoint of the segment $AC$ , and $M$ be the intersection point of the ray $KN$ and the circle.The tangents to the circle drawn at $A$ and $C$ meet at $E$ . prove that $\angle EMK = 90^o$

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#2 h Jul 21, 2020, 9:53 AM

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Solution

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#3 h Jul 22, 2020, 3:21 AM • 2 Y

Y by Mango247, Mango247

Let $O$ be the center of the circle. By symmetry, $E$ , $N$ , and $O$ are collinear. Also, $\angle ECO = \angle EAO = 90$ , so $ECOA$ is cyclic.
Then $MN \cdot NK = AN \cdot NC = EN \cdot NO$ , so $EMOK$ is cyclic. Then $\angle EMK = \angle EOK = \angle MOK = 90$ (since $M$ and $K$ are both midpoints of arcs $\widehat{AC}$ and $\widehat{BC}$ , respectively, and arc $\widehat{ACB}=180$ ).

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