如图，P为^ABC所在平面外一点，AP=AC,BP=BC,D为PC的中点 证明:直线PC与平面ABD垂直---**Extracted Content:** **Question 16 (10 points)** **Question Stem:** As shown in the figure, P is a point outside the plane containing △ABC. AP=AC, BP=BC, and D is the midpoint of PC. **Proof Required:** Prove that line PC is perpendicular to plane ABD. **Chart/Diagram Description:** * **Type:** Three-dimensional geometric figure, specifically a tetrahedron or pyramid with base ABC and apex P. * **Main Elements:** * **Points:** * Point P (top, apex) * Point A (bottom left) * Point B (bottom middle/right) * Point C (bottom right) * Point D (on line PC, between P and C) * **Lines/Edges:** * Solid lines: AB, BC, AC (forming the base triangle ABC); PA, PB, PC (edges connecting the apex P to the base vertices). * Dashed lines: AD, BD (lines within the plane ABD, likely representing lines that are obscured from view or constructed for the proof); PD (segment of PC, showing D as an interior point). * **Shapes:** * Triangle ABC (base) * Triangle PAB, PBC, PAC (side faces) * Triangle ABD * **Relative Position and Direction:** * Point P is shown above and connected to points A, B, C. * Points A, B, C appear to form a triangle on a lower plane. * Point D is located on the line segment PC. * The lines AD and BD are drawn, suggesting plane ABD. * The figure depicts spatial relationships for a 3D geometry problem.