5. Given a parallelogram $A B C D$. A circle $\Omega$ with a diameter of 13 is circumscribed around triangle $A B M$, where $M$ is the point of intersection of the diagonals of the given parallelogram. $\Omega$ intersects the ray $C B$ and the segment $A D$ again at points $E$ and $K$ respectively. The length of arc $A E$ is twice the length of arc $B M$ (arcs $A E$ and $B M$ do not have common points). The length of segment $M K$ is 5. Find the lengths of segments $A D, B K$ and the perimeter of triangle $E B M$.
Answer: $A D=13, B K=\frac{120}{13}, P_{E B M}=\frac{340}{13}$.

Solution. Let the degree measures of arcs $B M$ and $A E$ be $2 \alpha$ and $4 \alpha$ respectively. Then, by the inscribed angle theorem, $\angle A B E=\frac{1}{2} \cdot 4 \alpha=2 \alpha, \angle B A M=\frac{1}{2} \cdot 2 \alpha=\alpha$. Therefore, $\angle A B C=180^{\circ}-\angle A B E=180^{\circ}-2 \alpha ; \angle A C B=180^{\circ}-\angle B A C-\angle A B C=\alpha$.

In triangle $A B C$, the angles at vertices $A$ and $C$ are equal, so triangle $A B C$ is isosceles, $A B=B C$. Therefore, $A B C D$ is a rhombus. The diagonals of a rhombus are perpendicular to each other, so angle $A M B$ is a right angle and $A B$ is the diameter of the circle. Thus, $\angle B K A=\angle B E A=90^{\circ}$. Since $B E \| A K$, we have that $A E B K$ is a rectangle, and therefore, $E K$ is also a diameter. Finally, from the parallelism of $A D$ and $B C$, we have $\angle C A D=\angle A C B=\alpha$.

The diameter of the circle is 13, so $E K=A B=13 ; A D=A B=13$ (as sides of the rhombus). Chords $M K$ and $B M$ are equal (since the corresponding arcs are equal), $B M=5$. Then $A M=\sqrt{A B^{2}-B M^{2}}=12$, and from triangle $A B M$ we find that $\cos \alpha=\frac{A M}{A B}=\frac{12}{13}, \sin \alpha=\frac{B M}{A B}=\frac{5}{13}$. Therefore, $\sin 2 \alpha=2 \sin \alpha \cos \alpha=\frac{120}{169}, \cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=\frac{119}{169}$.

Next, we calculate: $B K=A B \sin 2 \alpha=13 \cdot \frac{120}{169}=\frac{120}{13} ; B E=A B \cos 2 \alpha=13 \cdot \frac{119}{169}=\frac{119}{13}$. Since $E K$ is a diameter, $\angle E M K=90^{\circ}, E M=\sqrt{E K^{2}-K M^{2}}=12$. Therefore, the perimeter of triangle $E B M$ is $5+12+\frac{119}{13}=\frac{340}{13}$.