解:过点M作ME‖AC,连结NE∵M是BG的中点∴ME是ΔBGC的中位线∴ME= E是BC中点∵N是CD中点∴EN是ΔCBD的中位线∴EN= ∵BD=CG∴ME=EN∴∠漏贺皮EMN=∠返差ENM∵ME‖AC∴∠EMN=∠AQP∵EN‖BD∴∠ENM=∠APQ∴∠APQ=∠AQP∴拍银AP=AQ