Estonian Math Competitions

As $AL\parallel BC$ and $AK\perp BC$ (Fig. 11), we have $\angle KAL=90^{\circ }$. As $KAL$ and $KCL$ are both right angles, points $A$ and $C$ lie on circle with diameter $KL$. By inscribed angles, $\angle KCA=\angle KLA$. On the other hand, $\angle KLA=\angle KBC$ and, by symmetry of isosceles triangle, $\angle KBC=\angle KCB$. Consequently, $\angle KCA=\angle KCB$, implying that $KC$ bisects the internal angle on vertex $C$ of the triangle $ABC$. As $KCL$ is right angle, $CL$ bisects the corresponding external angle.

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Alternative solution.

The altitude drawn from the vertex angle of the isosceles triangle $ABC$ is the perpendicular bisector of its base. All points of the perpendicular bisector of a line segment lie at equal distance from the endpoints of the line segment, implying that $|KB|=|KC|$. As $AL\parallel BC$ and $AK\perp BC$, we conclude $AK\perp AL$. Fig. 11 Let $P$ be a point on line $BC$ such that $C$ is between $B$ and $P$, and let lines $CK$ and $AL$ intersect at $Q$ (Fig. 12). As $\angle BKC=\angle LKQ$ and $\angle KBC=\angle KLQ$, triangles $KBC$ and $KLQ$ are similar. Together with the equality $|KB|=|KC|$, it implies $|KL|=|KQ|$. Thus $KA$ is the altitude drawn from the vertex angle of the isosceles triangle $KLQ$, implying also $|AL|=|AQ|$. Hence $A$ is the midpoint of the hypotenuse of the right triangle $CLQ$, implying that $A$ is the center of the circumcircle of the triangle $CLQ$. Consequently, $|AC|=|AL|$, implying that $\angle ACL=\angle ALC=\angle LCP$, i.e., $CL$ bisects the external angle on vertex $C$ of the triangle $ABC$. Fig. 12

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Alternative solution.

When point $K$ moves along the altitude drawn from vertex $A$ of the triangle $ABC$ farther from point $A$, the angle $KCB$ decreases, while point $L$ also moves farther from point $A$, causing the angle $LCB$ to increase. Thus the difference $\angle LCB-\angle KCB$ also increases in this process and can equal $90^{\circ }$ in the case of exactly one location of point $K$. Hence it suffices to show that if $L$ lies at the bisector of the external angle on vertex $C$ of the triangle $ABC$ then $\angle LCK=90^{\circ }$. So let $L$ lie at the bisector of the external angle on vertex $C$ of the triangle $ABC$ (Fig. 13). As $AL\parallel BC$ and $AK\perp BC$, we have $AK\perp AL$; as $AK$ is the altitude drawn from the vertex angle of the triangle $ABC$, it bisects the vertex angle, whence also $L$ lies on the bisector of the external angle on vertex $A$ of the triangle $ABC$. Thus $L$ is the center of the excircle tangent to side $AC$ of the triangle $ABC$ and lies also on the bisector of the internal angle on vertex $B$ of the triangle $ABC$. Hence both $AK$ and $BK$ are angle bisectors, implying that $K$ is the intersection point of angle bisectors of the triangle $ABC$ and $CK$ is the bisector of the internal angle on vertex $C$ of the triangle $ABC$. Consequently, $CK$ and $CL$ are perpendicular. Fig. 13