Through the vertices $A$ and $C$ of triangle $ABC$, a circle $K$ is drawn, the center of which lies on the circumcircle of triangle $ABC$. Circle $K$ intersects the extension of side $BA$ beyond point $A$ at point $M$.

Find the angle $C$, if $MA: AB=2: 5$, and $\angle B=\arcsin 3 / 5$.

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Let $A N$ be the height of triangle $A B C$. Prove that triangle $A N C$ is isosceles.

## Solution

Let $Q$ be the center of circle $K, A B=5 x, A M=2 x, A N$ be the height of triangle $A B C$. Then $A N=A B \sin \angle B=3 x$, $B N=4 x$,

$\angle A M C=1 / 2 \angle A Q C=1 / 2\left(180^{\circ}-\angle B\right)=90^{\circ}-1 / 2 \angle B$.

Since $Q A=Q C$, then $B Q$ is the bisector of angle $B$. Therefore, $\angle Q B M=1 / 2 \angle B$. Consequently, $M C \perp B Q$. Therefore, triangle $M B C$ is isosceles, $B M=B C=7 x$. Thus, $N C=B C-B N=7 x-4 x=3 x=N A$.

Therefore, $\angle C=\angle N C A=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-22.jpg?height=383&width=489&top_left_y=963&top_left_x=790)

Answer

$45^{\circ}$.