ABCD is a quadrilateral [Fig. 9.26]. A line through D, parallel to AC meets BC produced in P. Prove ar (△ABP) = ar (quad. ABCD).
Given: A quadrilateral ABCD in which DPII AC
To prove: ar (△ABP) = ar (quad. ABCD)
Proof: △ACP and △ACD are on same base AC and between same parallels AC and DP.
⇒ ar (△ACP) = ar (△ACD)
Adding, ar (△ABC) on both sides,
ar ( △ABC) + ar (△ACP) = ar (△ABC) + ar (ACD)
ar (△ABP) = ar (quad. ABCD)