7. In triangle $ABC$, the median $BM$ is drawn; $MD$ and $ME$ are the angle bisectors of triangles $AMB$ and $CMB$ respectively. Segments $BM$ and $DE$ intersect at point $P$, and $BP=1$, $MP=3$.

a) Find the segment $DE$.

b) Suppose it is additionally known that a circle can be circumscribed around quadrilateral $ADEC$. Find its radius.
Answer. a) $D E=6$; b) $R=3 \sqrt{65}$.

Solution. a) By the property of the angle bisector of a triangle, we get $A D: D B=A M: M B, C E: E B=C M: M B$, and since $A M=C M$, it follows that $A D: D B=C E: E B$, so $A C \| D E$. Then $\angle P D M=\angle A M D=\angle B M D$, which means that triangle $P D M$ is isosceles and $D P=M P=3$. Similarly, we get that $E P=3$ and then $D E=6$.

b) The trapezoid $A D E C$ is inscribed in a circle, so it is isosceles. The segment $P M$, connecting the midpoints of the bases of the trapezoid, is perpendicular to them. From the similarity of right triangles $B P D$ and $B M A$, we find that $M A=P D \cdot \frac{M P}{M B}=12$. Let $E H$ be the height of the trapezoid. Then $A H=A M+M H=A M+P E=15$, $A E=\sqrt{A H^{2}+H E^{2}}=3 \sqrt{26}, C H=C M-M H=9, C E=\sqrt{C H^{2}+H E^{2}}=3 \sqrt{10}, \sin \angle H C E=\frac{E H}{E C}=\frac{1}{\sqrt{10}}$. $3 \sqrt{65}$.

The given circle is the circumcircle of triangle $A C E$, so its radius $R$ is equal to $\frac{A E}{2 \sin \angle A C E}=$

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