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Answer» 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!
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