12.2. Consider the pyramid $V A B C$, where $A B=7 \, \text{cm}, A C=5 \, \text{cm}, B C=8 \, \text{cm}, V B=6 \, \text{cm}$, and the dihedral angles at the base $A B C$ are congruent. Determine the distance from point $P$ to the edge $V A$, where $A P$ is the angle bisector in triangle $A B C$.
Solution. The projection of vertex $V$ onto the base $A B C$ is the center $O$ of the circle inscribed in triangle $A B C$. Let $M$ and $N$ be the points of tangency of the sides $A B$ and $B C$, respectively, with the circle inscribed in triangle $A B C$. Using Heron's formula, we calculate $\mathcal{A}_{\triangle A B C}=10 \sqrt{3} \mathrm{~cm}^{2}$. Then $O M=O N=\sqrt{3} \mathrm{~cm}$.

Applying the theorem about tangents drawn from an external point to a circle, we get $B N=5 \mathrm{~cm}, A M=2 \mathrm{~cm}$.

Then $O B=2 \sqrt{7} \mathrm{~cm}, A O=\sqrt{7} \mathrm{~cm}$.

We calculate $V O=2 \sqrt{2} \mathrm{~cm}$ and $V A=\sqrt{15} \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_aa53cb7a74c6eadc2c82g-1.jpg?height=458&width=393&top_left_y=1950&top_left_x=1430)

Let $A P$ be the angle bisector of $\angle A$. According to the angle bisector theorem, we get $P B=\frac{14}{3} \mathrm{~cm}$. Then $N P=\frac{1}{3} \mathrm{~cm}, O P=\frac{2 \sqrt{7}}{3} \mathrm{~cm}$,

$A P=\frac{5 \sqrt{7}}{3} \mathrm{~cm}$ and $V P=\frac{10}{3} \mathrm{~cm}$. Let $d$ be the distance from point $P$ to the edge $V A$.

Then $\mathcal{A}_{A P V}=\frac{1}{2} A P \cdot O V=\frac{1}{2} V A \cdot d$ implies $d=\frac{2 \sqrt{210}}{9} \mathrm{~cm}$.