如图，P是等边三角形ABC内的一点，且PA = 3，PB = 4，PC = 5，以BC为边在△ABC外作△BQC≌△BPA，连接PQ，则以下结论中正确有（$\qquad$） （填序号） ①△BPQ是等边三角形 ②△PCQ是直角三角形 ③∠APB = 150° ④∠APC = 135°---**Diagram Description:** * **Type:** Geometric figure (3D, likely a tetrahedron or pyramid with internal points). * **Main Elements:** * **Points:** Five labeled points: A, B, C, P, Q. * **Lines:** Straight lines connecting the points: AB, AC, BC, BP, CP, AP, AQ, PQ. These represent edges or segments within the figure. * **Shapes:** Triangle ABC forms the base or one face. Q is a point on the edge BC. P is an interior point within the region formed by vertices A, B, C, or possibly on the line segment AQ. The figure resembles a tetrahedron with vertices A, B, C, and a point P inside, and a point Q on edge BC. * **Labels and Annotations:** The vertices and two interior/edge points are labeled as A, B, C, P, and Q. * **Relative Position and Direction:** A is depicted as an apex. B and C form part of a base or face. Q is shown on the line segment BC. P is shown inside the solid formed by A, B, C, or on the line segment AQ, and is connected to A, B, and C.