Question

In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that ar(∆ABP) = ar(quad. ABCD).

Answer 1

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)​
Adding ar(∆ABC) on both sides, we get:
⇒ ar(∆ACD) + ar(∆ABC) = ar(∆ACP)​​ + ar(∆ABC)
⇒​ ar(quad. ABCD) = ar(∆ABP)
​Hence, proved.