帮我讲解这道几何数学题---**Textual Information:** 如图, 分别以△ABC 的 AC 和 BC 为一边, 在△ABC 的外侧作正方形 ACDE 和正方形 CBFG, 点 P 是 EF 的中点. 求证: 点 P 到边 AB 的距离等于 AB 的一半. **Chart/Diagram Description:** * **Type:** Geometric figure illustrating a proof problem. * **Main Elements:** * **Points:** A, B, C, D, E, F, G, P, Q. * **Shapes:** * Triangle ABC. * Square ACDE constructed externally on side AC. * Square CBFG constructed externally on side BC. * **Lines:** Segments connecting the points as labeled (AC, BC, AB, CD, DE, EA, CB, BF, FG, GC, EF, AP, BP, CP, EP, FP, PQ). PQ is drawn perpendicular to AB. * **Relative Position and Direction:** * Squares ACDE and CBFG are constructed on the outside of triangle ABC. * Point P is the midpoint of the segment EF. * Point Q is on the line segment AB, such that PQ is perpendicular to AB. * **Labels and Annotations:** Points are labeled with letters A, B, C, D, E, F, G, P, Q. The line segment PQ is depicted and labeled, indicating the distance from P to AB.