A3. Using the angles of triangle $A B M$, we calculate $\Varangle M B A=180^{\circ}-84^{\circ}-48^{\circ}=48^{\circ}$, hence triangle $A B M$ is isosceles with the vertex at $A$. It follows that $|A B|=|A M|$. Since $M$ is the midpoint of side $A D$, it holds that $|A M|=|M D|$, and since $A B C D$ is a parallelogram, $|A B|=|C D|$. Therefore, $|C D|=|M D|$ and triangle $M C D$ is also isosceles with the vertex at $D$. From this, we calculate $\Varangle D C M=\frac{180^{\circ}-\Varangle M D C}{2}=\frac{\Varangle B A D}{2}=42^{\circ}$.
A3. Point $M$ is the midpoint of side $A D$ of parallelogram $A B C D$ (see figure). Angle $B A D$ is $84^{\circ}$, and angle $A M B$ is $48^{\circ}$. How many degrees is angle $D C M$?
(A) 36
(B) 42
(C) 44
(D) 45
(E) 48

![](https://cdn.mathpix.com/cropped/2024_06_07_a0d1506e8e2028d6d427g-10.jpg?height=311&width=531&top_left_y=1675&top_left_x=1365)