Fig. 4

Since $K$ lies on the perpendicular bisector of the segment $AO$ (Fig. 4) we have $KA = KO$. On the other hand $KO = AO$ since $K$ and $A$ are points on the circle. Hence $AKO$ is an equilateral triangle from which we obtain that $\angle AOK = 60^\circ$. Hence $\angle KOB = \angle AOB - \angle AOK = 90^\circ - 60^\circ = 30^\circ$. As also $B$ lies on the same circle we have $KO = BO$, hence $\angle BKO = \frac{180^\circ - \angle KOB}{2} = \frac{180^\circ - 30^\circ}{2} = 75^\circ$. Finally from the equality $AO = BO$ we obtain $\angle ABO = \frac{180^\circ - \angle AOB}{2} = \frac{90^\circ}{2} = 45^\circ$. Hence $\angle BLK = \angle LOB + \angle LBO = 30^\circ + 45^\circ = 75^\circ$. The triangle $KBL$ is isosceles as it has two equal angles.