Open Contests

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.