In trapezoid $A B C D A D \| B C$, point $K$ is located on diagonal $B D$, and $B K: K D=1: 2$. Find the angles of triangle $A K C$, if $A C=A D-2 B C, \angle C A D=\alpha$.

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Let $P$ be a point on the base $A D$ such that $D P=2 B C$. Then the triangle $C A P$ is isosceles, and the segment $C P$ divides the diagonal $B D$ in the ratio $1: 2$.

## Solution

Extend the base $B C$ beyond point $C$ and lay off a segment $C M$ equal to $B C$, and construct the parallelogram

$B M D P$. Then $D P=2 B C, A P=A C=c$. Therefore, the triangle $A C P$ is isosceles, $\angle A C P=\pi-\alpha / 2$.

The point of intersection of segments $C P$ and $B D$ divides these segments in the ratio $1: 2$, that is, it coincides with point $K$.

Drop perpendiculars $P L$ and $K N$ to $A C$. Then $K N=1 / 3 P L=c / 3 \sin \alpha$, and $A N=(A C-C N)=(c-1 / 3 C L)=(c-c / 3$

$(1-\cos \alpha))=c / 3(2+\cos \alpha)$. Therefore, $\operatorname{tg} \angle C A K=\operatorname{tg} \angle M A K=\frac{\sin \alpha}{2+\cos \alpha}$.

Answer

$\frac{\pi-\alpha}{2}, \operatorname{arctg}\left(\frac{\sin \alpha}{2+\cos \alpha}\right), \frac{\pi+\alpha}{2}-\operatorname{arctg}\left(\frac{\sin \alpha}{2+\cos \alpha}\right)$.