In the quadrilateral $A B C D$ (with internal angles $\alpha, \beta, \gamma, \delta$ it holds: $\alpha=\gamma=90^{\circ},|A D|=|C D|$ and $\beta=\delta+100^{\circ}$. Let $M$ be the intersection of the angle bisectors of $\angle D A C$ and $\angle A C D$. What is the measure of the internal angle at vertex $M$ in triangle $A M D$.

(Ptáčková)
First, we will calculate the sizes of the remaining angles. The sum of the interior angles in any quadrilateral is equal to $360^{\circ}$, so $\alpha+\beta+\gamma+\delta=360$. Substituting the given values, we get $90+(\delta+100)+90+\delta=360$, from which $\delta=40^{\circ}$ and $\beta=140^{\circ}$.

Now we will focus only on triangle $A C D$ (see figure). It is isosceles with base $A C$, as stated in the problem. We will determine the size of angle $\omega$ at the base. It holds that $180^{\circ}=40^{\circ}+2 \omega$, so the size of the angle at the base is $70^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5b7c063dbc9d94c6df02g-1.jpg?height=409&width=551&top_left_y=1640&top_left_x=792)

$\triangle A M D$: the interior angle at vertex $A$ is $70^{\circ}: 2=35^{\circ}$, the interior angle at vertex $D$ is $40^{\circ}: 2=20^{\circ}$. The size of the interior angle at vertex $M$ is $180^{\circ}-35^{\circ}-20^{\circ}=125^{\circ}$.