Question

solve the question please

Answer 1

Hello Mate!

(a) Since ∆APD and ||gm ABCD lie on same base and same parallels so,

ar(∆APD) = ½ ar(||gmABCD) __(i)

ar(∆APD) + ar(∆DCP) + ar(∆APB) = ar(||gmABCD)

½ ar(||gmABCD) + ar(∆DCP) + ar(∆APB) = ar(||gmABCD)

ar(∆DCP) + ar(∆APB) = ar(||gmABCD) - ½ ar(||gmABCD)

ar(∆DCP) + ar(∆APB) = ½ ar(||gmABCD) __(ii)

From (i) and (ii) we get,

ar(∆APD) = ar(∆DCP) + ar(∆APB)

(b) Through O draw EF || BC and GH || AB.

Hence, we get ABHG and GHCD are two parallelogram.

ar(∆AOB) = ½ ar(||gmABHG)

ar(∆COD) = ½ ar(||gmGHCD)

Adding 2 equations we get,

ar(∆AOB) + ar(∆COD) = ½ ar(||gmABHG) + ½ ar(||gmGHCD)

ar(∆OAB) + ar(∆OCD) = ½ ar(||gmABCD)

Similarly we can prove,

ar(∆OBC) + ar(∆OAD) = ½ ar(||gmABCD)

Hence proved, Q.E.D

Have great future ahead!