Question

In the adjoining figure,ABCD is a quadrilateral.A line through D,parallel to AC,meets BC produced in P.

Prove that $ar\left(△ABP\right)=ar(quad.ABCD)$.

Answer 1

ANSWER:
We have:
ar(quad. ABCD) = ar(∆ ACD) + ar(∆ ABC)
ar(∆ ABP ) = ar(∆ ACP)​​ + ar(∆ ABC)
∆ ACD and ∆ ACP are on the same base and between the same parallels AC and DP.
∴ ar(∆ ACD) = ar(∆ ACP)
By adding ar(∆ ABC) on both sides, we get:
ar(∆ ACD) + ar(∆ ABC) = ar(∆ ACP)​​ + ar(∆ ABC)
⇒​ ar (quad. ABCD) = ar(∆ ABP )
Hence, proved.