Note that $\angle ADB = 90^\circ = \angle ACB$ (angles in a semicircle). It follows that $E$ is the orthocentre of $\triangle FAB$. Therefore $FK$ is perpendicular to $AB$. It follows that $ADEK$ is cyclic. Therefore $\angle AEK = \angle ADK$. Since $\angle ADK = \angle ADL = \angle ACL$ (same segment), we have $\angle AEK = \angle ACL$ which implies that $EK$ is parallel to $CL$. Since $EK$ is perpendicular to $AB$, this implies that $CL$ is perpendicular to $AB$.

Because the angles $\angle ADB$ and $\angle ACB$ are right angles, $E$ is the orthocentre of triangle $ABF$ and $FK$ is perpendicular to $AB$. From the right angles $\angle BKF$ and $\angle BDF$ we see that $BFDK$ is cyclic, hence $\angle KDB = \angle KFB$. Because $ABCD$ was given to be cyclic, we have $\angle LAB = \angle LDB$. The right angled triangles $ABC$ and $FKB$ are similar as they share an angle at $B$, and so $\angle KFB = \angle BAC$. Together, this implies
$$\angle LAB = \angle LDB = \angle KDB = \angle KFB = \angle BAC$$
and this means that $AB$ is the angle bisector of $\angle LAC$.

As a consequence, the right angled triangles ALB and ACB are congruent, which shows that triangle LAC is isosceles with |AL| = |AC|. It is a well known fact about isosceles triangles, which easily follows from ASA, that the angle bisector AB is perpendicular to the opposite side CL.

It is known that the altitudes of a triangle are the internal angle bisectors of its orthic triangle. Armed with this knowledge we proceed as follows.
Because the angles $\angle ADB$ and $\angle ACB$ are right angles, $E$ is the orthocentre of triangle $ABF$ and $FK$ is perpendicular to $AB$. Hence, $CDK$ is the orthic triangle of triangle $ABF$, which implies that $BD$ is the angle bisector of $\angle KDC = \angle LDC$.

Now we use the fact that an angle bisector in a triangle meets the perpendicular bisector of the opposite side on the circumcircle of that triangle. Applying this to triangle LDC we see that B is the point on the circumcircle where the angle bisector of $\angle LDC$ meets the perpendicular bisector of CL, which passes through the circumcentre O of triangle LDC. Hence, BO is perpendicular to CL.