60th Belarusian Mathematical Olympiad

Let $P$ and $M$ be marked on $AD$ and $BC$ so that $MP\perp AD$ and $BM=MC$. By condition, $K$ belongs to $MP$. Let $L$ and $N$ be respectively the feet of the perpendiculars from $K$ to the lines $AB$ and $DC$. Two cases are possible:

1) Both the points lie on the sides $AB$ and $DC$ (or on their extensions).

2) One of the points lies on the side and the other point lies on the extension of the other side.

Since $AK$ and $DK$ are bisectors, we have $KL=KP=KN$. Since $K$ lies on the perpendicular bisector, we have $BK=KC$. Therefore the right-angled triangles $BLK$ and $CNK$ are equal, so $\angle LBK=\angle NCK$, $BL=CN$.

For the first case (see Fig. 1) from the isosceles triangle $BKC$ it follows that $\angle KBC=\angle KCB$. So $$\begin{array}{rl}\angle ABC& =\angle ABK+\angle KBC=(180^{\circ }-\angle LBK)+\angle KBC=\\ & =(180^{\circ }-\angle NCK)+\angle KCB=\angle KCD+\angle KCB=\angle BCD.\end{array}$$ Therefore, $\angle ABC=\angle DCB$, and so the trapezium $ABCD$ is isosceles. (Similarly we can consider the cases when both $L$ and $N$ lie on the sides $AB$ and $DC$).

Fig. 1

If we have the second case (see Fig. 2), then $AL=AP$, $PD=DN$ (since $AK$ and $DK$ are bisectors). Therefore, $$AB+DC=(AL+LB)+(DN-NC)=AP+PD+(LB-NC)=AP+PD=AD,$$ as required.

Fig. 2