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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

By midpoint theorem HG // AC and HG = 1/2 AC eq1 Similarly EF // AC EF = 1/2 AC eq2 From eq 1 and eq 2 HG // EF and HG =EF therefore EFGH is a parallelogram

In ΔHEF and //gm HABF lie on the same base HF and between same p//s HF and AB therefore ar(ΔHEF) = 1/2 ar(HABF) eq3 similarly ar(ΔHGF) = 1/2 ar(HFCD) eq4 adding eq3 and eq 4 ar (ΔHEF) + ar(ΔHGF) = 1/2 ar (HABF+HFCD) ar(EFGH) = 1/2 ar (ABCD)