Question 32
Let $ABC$ be an acute-angled triangle and $D$ be the foot of the altitude from $A$. Let the tangent line at $D$ to the circle with diameter $AB$ meet the circle with diameter $AC$ again at $K$, and let the tangent line at $D$ to the circle with diameter $AC$ meet the circle with diameter $AB$ again at $L$. Point $S$ is given on the plane such that
$$
\angle ABC + \angle ABS = \angle ACB + \angle ACS = 180^\circ.
$$
Prove that $A$, $K$, $L$ and $S$ lie on a circle.
$$
\angle ABC + \angle ABS = \angle ACB + \angle ACS = 180^\circ.
$$
Prove that $A$, $K$, $L$ and $S$ lie on a circle.