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Given- two llgms ABCD and ABFE ON THE SAME BASE and b/w the same parallels DF and AB proof-in triangle ADE and triangle BCF angle 4= angle 3 ( corresponding angles) angle 1 =angle 2(same) AD=BC(OPP.SIDES OF A llgm) therefore , triangle ADE IS CONGRUENT TO triangle ABC by ASA so triangle ADE= triangle BCF THEREFORE ar (ade) =ar(bcf)(congruent fig.have =areas) now ar(ABCD)=AR (ADE)+ar(EDCB) =ar(BCF)+ar(EDCB) =ar(EFCD) SO ,llgms ABCD =EFCD are equal in areas