Question 88
Two points $M$ and $N$ have been selected from the edge $BC$ of a tetrahedron $ABCD$ so that the point $M$ lies between $B$ and $N$, the point $N$ lies between $M$ and $C$, and the angle between the planes $AND$ and $ACD$ is equal to the angle between the planes $AMD$ and $ABD$. Prove that
$$
\frac{MB}{MC} + \frac{NB}{NC} \ge 2 \cdot \frac{|\triangle ABD|}{|\triangle ACD|},
$$
where $|\triangle ABD|$ is the area of the triangle $\triangle ABD$, and similarly $|\triangle ACD|$ is the area of $\triangle ACD$.
$$
\frac{MB}{MC} + \frac{NB}{NC} \ge 2 \cdot \frac{|\triangle ABD|}{|\triangle ACD|},
$$
where $|\triangle ABD|$ is the area of the triangle $\triangle ABD$, and similarly $|\triangle ACD|$ is the area of $\triangle ACD$.