5. In triangle $\mathrm{ABC}$, point $\mathrm{M}$ is the midpoint of $\mathrm{AC}$, $\mathrm{MD}$ and $\mathrm{ME}$ are the angle bisectors of triangles $\mathrm{ABM}$ and $\mathrm{CBM}$, respectively. Segments $\mathrm{BM}$ and $\mathrm{DE}$ intersect at point $\mathrm{F}$. Find $\mathrm{MF}$, if $\mathrm{DE}=7$.
# Answer: 3.5

Solution. By the property of the angle bisector from triangles AMB and CMB, we get that $\frac{A D}{B D}=\frac{A M}{B M}$ and $\frac{C E}{B E}=\frac{C M}{B M}$. According to the condition, $A M=C M$, therefore, $\frac{A D}{B D}=\frac{C E}{B E}$, hence, $D E \| A C$ (by the converse of Thales' theorem for angle $A B C$ or from the similarity of triangles $D B E$ and $A B C$). Then $F$ is the midpoint of segment $D E$.

Since MD and ME are the bisectors of adjacent angles, triangle DME is a right triangle. Its median MF, drawn from the vertex of the right angle, is equal to half the hypotenuse $D E$.