帮我解析这道题---**Extracted Content:** **Question 16 (10 points)** **Question Stem:** As shown in the figure, P is a point outside the plane containing triangle ABC. Given AP = AC, BP = BC, and D is the midpoint of PC. **Proof Requirement:** Prove that line PC is perpendicular to plane ABD. **Chart/Diagram Description:** * **Type:** Geometric figure, specifically a 3D polyhedral representation (likely a tetrahedron or pyramid with vertices A, B, C, P), projected into 2D. * **Main Elements:** * **Points:** Five labeled points are visible: A (bottom left), B (bottom right), C (middle right), P (top left), and D (on the line segment PC, roughly in the middle). * **Lines/Segments:** * Solid lines connect A, B, and C, forming triangle ABC. * Dashed lines represent edges that would be hidden from view: PA, PB, PC. * Dashed lines also connect point D to A and B, forming segments AD and BD. * Overall, the figure depicts a triangular pyramid P-ABC, with additional lines AD and BD forming plane ABD, and point D being the midpoint of edge PC. * **Relative Position and Direction:** * Points A, B, C form a triangle that appears to be the base. * Point P is positioned above and to the left of the base triangle. * Point D is located on the segment PC, indicating it's the midpoint of PC as stated in the text. * The lines PA, PB, PC, AB, AC, BC, AD, BD are drawn to connect the points.