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ABCD is a rectangle and EFGH is a trapezium in the given figure. Prove that ar(ABCD) = ar(EFGH)

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Given: ABCD and EFGH are rectangle and trapezium, respectively, DE=CH and AH||FC.To prove: arABCD=arEFGHProof:∵∆DEF and ∆CHG has equal bases DE and CH; and are between same parallel lines FG and AH.∴ar∆DEF=ar∆CHG        .....iAs, AH||FC , AF||QC and FP||CHSo, AFCQ and FCHP are parallelograms which are on same base FC and are between same parallels AH and FC∴arAFCQ=arFCHP        .....iiSimilarly, rectangle ABGF and parallelogram AFCQ are on same base AF and are between same parallels AF and BCSo, arABGF=arAFCQ        .....iiiAlso, ∆CHG and parallelogram FCHP are on same base CH and are between same parallels CH and FGSo, ar∆CHG=12arFCHP        .....ivFrom ii,iii and iv, we getar∆CHG=12arABGF⇒arABGF=2 ar∆CHG⇒arABGF=ar∆CHG+ar∆DEF      From i          .....vNow,arABCD=arABGF+arDCGF=ar∆CHG+ar∆DEF+arDCGF        From v∴arABCD=arEFGH

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