10 Unfold the triangular pyramid $P-ABC$ along its three lateral edges, forming a planar shape as shown in the figure. Points $P_{1}, C, P_{2}$ are collinear, points $P_{2}, B, P_{3}$ are collinear, and $P_{1} P_{2}=P_{2} P_{3}$. Then, in the triangular pyramid $P-ABC$, the angle formed by $PA$ and $BC$ is ( ).
(Problem 10)
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
10 In the development diagram, point $C$ is the midpoint of line segment $P_{1} P_{2}$, $B$ is the midpoint of line segment $P_{2} P_{3}$, and $P_{1} A = P_{3} A$, so $\angle A P_{1} P_{3} = \angle P_{1} P_{3} A$. From $P_{1} P_{2} = P_{2} P_{3}$, we know
$$
\begin{array}{c}
\angle P_{2} P_{1} P_{3} = \angle P_{1} P_{3} P_{2}, \\
P_{1} C = P_{3} B = P_{2} C = P_{2} B,
\end{array}
$$

Therefore,
$$
\triangle A P_{1} C \cong \triangle A P_{3} B, A C = A B .
$$

In the triangular pyramid, take the midpoint $D$ of $B C$, and connect $P D$ and $A D$, then $P D \perp B C, A D \perp B C$, so $B C \perp$ plane $P A D, P A \perp B C$. Hence, the correct choice is D.