6. In a convex quadrilateral $A B C D$, the bisector of angle $B$ passes through the midpoint of side $A D$, and $\angle C=\angle A+\angle D$. Find the angle $A C D$. (S. Berlov)
Answer. $\angle A C D=90^{\circ}$. Solution. Let $E$ be the midpoint of side $A D$, and $F$ be the

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intersection point of $B E$ and $A C$. From the given condition, we have: $\angle B=360^{\circ}-2(\angle A+\angle D)$, from which $\angle A E B=180^{\circ}-\angle A-$ $\angle B / 2=\angle D$. Therefore, $B E \| C D$, and $E F$ is the midline of triangle $A C D$, meaning $A F=F C$. Thus, $B F$ is the bisector and median of triangle $A B C$, and therefore its altitude. Consequently, line $C D$, parallel to $B F$, is also perpendicular to $A C$, from which the answer follows.