In the quadrilateral $A B C D$, $A B=C D, A B C \angle=90^{\circ}$ and $B C D \angle=100^{\circ}$. The intersection point of the perpendicular bisectors of sides $A D$ and $B C$ is $M$. Determine the angle $B M C$.

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Let $E$ be the point such that the quadrilateral $A B C E$ is a rectangle. Then $A B=C E=D C$ and $\angle D C E=\angle D C B-90^{\circ}=10^{\circ}$ (Figure 1). The perpendicular bisector of segment $A E$ coincides with the perpendicular bisector of $B C$, so $M$ is the center of the circumcircle of triangle $A E D$. Therefore, the perpendicular bisector of $E D$ also passes through $M$. However, triangle $E D C$ is isosceles, so the perpendicular bisector of $E D$ is also the angle bisector of $\angle E C D$, which forms a $5^{\circ}$ angle with $E C$.

Let $F$ be the midpoint of $B C$ and $G$ be the midpoint of $E D$. Then $\angle C M F=\angle G C E=5^{\circ}$, because $C E \parallel M F$; $G$, $C$, and $M$ lie on the same line. Triangle $C M B$ is also isosceles, so $\angle C M F=\angle F M B$, which means $\angle B M C=\angle B M F+\angle F M C=10^{\circ}$.

Based on the work of Gergely Jánosi (Budapest, Eötvös J. Gymnasium, 1st year)

Note. Many made the following mistake: "Using the notation of Figure 2: $A M=M D, B M=M C$ and $A B=C D$, so triangles $A B M$ and $D C M$ are congruent, which means $\angle A B M=\angle D C M$. Furthermore, triangle $B M C$ is isosceles, so $\angle M B C=\angle M C B$. From these, it follows that

$90^{\circ}=\angle A B C+\angle M B C=\angle D C M+\angle M C B=$

$$
=\angle D C B=100^{\circ}
$$

This is a contradiction."

This reasoning is incorrect because the points are not actually arranged as shown in the diagram. The sketch is not to scale, and $D C < A B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_810efd21623cbd748640g-1.jpg?height=1008&width=1225&top_left_y=976&top_left_x=428)