证明:分别过P作AC、AB的平行线交AB、BD于M、N,分别过P作AB、BD的垂线垂足为H、K,作点N关于PK的对称点N',连接PN',可证MBNP为平行四边形,∴∠PMB=∠PNB,PM/MB=AC/AB,∴PM:PN=AC:AB=PQ:PD,△PMH∽△PN'K,∴PM:PH=PN':PK,∴PM:PN=PM:PN'=PH:PK=PQ:PD,∴PH:PQ=PK:PD,∴sin∠Q=sin∠PDK,∵∠Q<∠ABP=75°,∠PDB<∠CDB=60°,∴∠Q=∠PDB,∴△PMQ∽△PN'D,∴∠MPQ=∠N'PD,∴∠MPQ-∠N'PQ=∠N'PD-∠N'PQ,即∠QPD=∠MPN'=∠PN'D=∠PMQ=∠BAC,证毕
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