Through the vertex $A$ of an acute-angled triangle $ABC$, a line parallel to the side $BC$, equal to $a$, is drawn and intersects the circles constructed on the sides $AB$ and $AC$ as diameters, at points $M$ and $N$, different from $A$. Find $MN$.
$\angle A M B=\angle A N C=90^{\circ}$.

## Solution

Since point $M$ lies on the circle with diameter $A B$, then $\angle A M B=90^{\circ}$. Similarly, $\angle A N C=90^{\circ}$. Therefore, the opposite sides of quadrilateral $B M N C$ are parallel in pairs. Consequently, $M N=B C=a$.

Answer

## a.