Since CQ || PD, triangles DPQ and DPC share base PD and parallel lines, giving Area(ΔDPQ) = Area(ΔDPC). Thus, Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC) = (1/2) x Area(ΔABC).
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Since CQ || PD, triangles DPQ and DPC share base PD and parallel lines, giving Area(ΔDPQ) = Area(ΔDPC). Thus, Area(ΔBPQ) = Area(ΔBPD) + Area(ΔDPQ) = Area(ΔBPD) + Area(ΔDPC) = Area(ΔBDC) = (1/2) x Area(ΔABC).