$91 P$ is inside the tetrahedron $A B C D$. If the distances from $P$ to the faces $B C D, A C D, A B D, A B C$ are $a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}$ respectively, and the distances from the vertices $A, B, C, D$ to the respective faces $B C D, A C D, A B D, A B C$ are $a, b, c, d$, then $\frac{a^{\prime}}{a}+\frac{b^{\prime}}{b}+\frac{c^{\prime}}{c}+\frac{d^{\prime}}{d}$ is
A. 1
B. $\frac{3}{2}$
C. 2
D. $\frac{1}{2}$
91 A. $\frac{a^{\prime}}{a}=\frac{V_{P B C D}}{V_{A B C D}}$ etc., so $\frac{a^{\prime}}{a}+\frac{b^{\prime}}{b}+\frac{c^{\prime}}{c}+\frac{d^{\prime}}{d}=1$.