3. In triangle $ABC$, two altitudes $AK$ and $CL$ are drawn. Find the measure of angle $B$, given that $AC = 2 \cdot LK$.
Solution. Construct a circle on side $AC$ as its diameter, which will pass through points $L$ and $K$, since $\angle ALC = \angle AKC = 90^{\circ}$. According to the condition, $AC = 2 \cdot LK$, and thus, the segment $LK$ is equal to the radius of the constructed circle, so the arc subtended by the chord $LK$ is $60^{\circ}$. Therefore, the angle $\angle LCK$, which subtends this arc, is $30^{\circ}$. Next, if angle $B$ is acute, then $\angle B = 90^{\circ} - \angle LCB = 60^{\circ}$ (Fig. $2, a$); if angle $B$ is obtuse, then $\angle CBL = 90^{\circ} - \angle BCL = 60^{\circ}$ (Fig. 2, b), and thus, $\angle B = 180^{\circ} - \angle CBL = 120^{\circ}$.
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