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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)
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)
SO ,llgms ABCD =EFCD are equal in areas

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