8,9  On the hypotenuse $BC$ of the right triangle $ABC$, points $D$ and $E$ are marked such that $AD \perp BC$ and $AD = DE$. On the side $AC$, a point $F$ is marked such that $EF \perp BC$. Find the angle $ABF$.}
Points $A$ and $E$ lie on a circle with diameter $B F$. The inscribed angles $A B F$ and $A E F$ are equal. In the isosceles right triangle $A D E$, the angles at the base are $45^{\circ}$ each, so $\angle A B F=\angle A E F$ $=90^{\circ}-\angle A E D=45^{\circ}$.

## Answer

$45^{\circ}$.