Problem 10.4. Given a parallelogram $A B C D$, point $M$ is the midpoint of side $B C$. On side $A D$, there is a point $K$ such that $B K=B M$ and the quadrilateral $K B M D$ is cyclic.

(a) (2 points) What is the length of segment $M D$, if $A D=17$?

(b) (2 points) How many degrees does the angle $K M D$ measure, if $\angle B A D=46^{\circ}$?

![](https://cdn.mathpix.com/cropped/2024_05_06_baaf26a33e2d9a36d9b4g-03.jpg?height=278&width=505&top_left_y=301&top_left_x=468)
Answer: (a) 8.5. (b) 48.

Solution. (a) Note that KBMD is a cyclic trapezoid, so it is isosceles, i.e., $M D=K B$. Therefore,

$$
M D=K B=\frac{B C}{2}=\frac{A D}{2}=8.5
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_baaf26a33e2d9a36d9b4g-03.jpg?height=361&width=616&top_left_y=877&top_left_x=419)

Fig. 7: to the solution of problem 10.4

(b) Since $M D=M B=M C$, angle $B D C$ is a right angle (as is known, if in a triangle the median is equal to half the side to which it is drawn, then it is drawn from a right angle). From the fact that $A B \| C D$, it follows that angle $A B D$ is also a right angle (Fig. 7).

Thus, $\angle B D K=90^{\circ}-\angle B A D$, and from the cyclic nature of quadrilateral $K B M D$ it follows that $\angle B M K=\angle B D K=90^{\circ}-\angle B A D$.

Moreover, from the parallelism it follows that $\angle M B D=\angle B D K$, and from the isosceles nature of triangle $B M D$ that $\angle B M D=180^{\circ}-2 \angle M B D=2 \angle B A D$.

We obtain

$$
\begin{aligned}
\angle K M D & =\angle B M D-\angle B M K=2 \angle B A D-\left(90^{\circ}-\angle B A D\right)= \\
& =3 \angle B A D-90^{\circ}=3 \cdot 46^{\circ}-90^{\circ}=48^{\circ} .
\end{aligned}
$$