Since ΔABQ and ΔPBQ lie on the same base BQ and are between the same parallels AP and BQ,
⊥ Area (ΔABQ) = Area (ΔPBQ) ... (1)
Again, ΔBCQ and ΔBRQ lie on the same base BQ and are between the same parallels BQ and CR.
⊥ Area (ΔBCQ) = Area (ΔBRQ) ... (2)
On adding equations (1) and (2), we obtain
Area (ΔABQ) + Area (ΔBCQ) = Area (ΔPBQ) + Area (ΔBRQ)
⊥ Area (ΔAQC) = Area (ΔPBR)