Through the midpoint of the hypotenuse of a right triangle, a perpendicular is drawn to it. The segment of this perpendicular, enclosed within the triangle, is equal to c, and the segment enclosed between one leg and the extension of the other is equal to 3c. Find the hypotenuse.

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Let $M$ be the midpoint of the hypotenuse $AB$ of the right triangle $ABC$, $N$ and $K$ be the points of intersection of the perpendicular to $AB$ passing through point $M$ with the leg $AC$ and the extension of the leg $BC$ respectively. Denote $AM = BM = x$. Due to the obvious similarity of triangles $AMN$ and $KMB$,

$\angle BKM = \angle BAC = \alpha$, so $MN : AM = BM : MK$, or $c / x = x / 4c$, from which $x = 2c$. Therefore, $AB = 2x = 4c$.

## Answer

$4c$.