4. In quadrilateral $A B C D$, points $X, Y, Z$ are the midpoints of segments $A B, A D, B C$ respectively. It is known that $X Y$ is perpendicular to $A B$, $Y Z$ is perpendicular to $B C$, and the measure of angle $A B C$ is $100^{\circ}$. Find the measure of angle $A C D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43d6a72f2e2cc851afdg-2.jpg?height=426&width=623&top_left_y=1392&top_left_x=725)
Solution 1. Triangles $A Y B, B Y C, A Y C$ and $C Y D$ are isosceles, as follows from the condition. Let the equal angles be $\angle B A Y=\angle A B Y=x, \angle C B Y=\angle B C Y=y$, $\angle D C Y=\angle C D Y=z, \angle C A Y=\angle A C Y=v$ (see fig.). We are given that $x+y=100^{\circ}$. We need to find $z+v$.

The sum of the angles in quadrilateral $A B C D$ is $x+x+y+y+z+z=360^{\circ}$. From this, $z=80^{\circ}$. Let's calculate the angle at the vertex of the isosceles triangle $C Y D$. $\angle C Y D=180^{\circ}-2 z=20^{\circ}$. This angle is the external angle of triangle $A Y C$, so $\angle C Y D=2 v$, from which $v=10^{\circ}$. Now we can find the measure of angle $A C D$. $\angle A C D=z+v=80^{\circ}+10^{\circ}=90^{\circ}$.

Solution 2. From the condition, it immediately follows that triangles $A Y B$ and $B Y C$ are isosceles, and the lateral sides are equal to each other and equal to segments $A Y$ and $Y D$. We get that point $Y$ is the center of the circle circumscribed around quadrilateral $A B C D$, and $A D$ is the diameter of this circle. Therefore, angle $A C D$ is a right angle. Answer. $\angle A C D=90^{\circ}$.