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Question
The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF draw parallel to DB meets AB produced at F.
Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.
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Solution
Since triangle EDG and EGA are on the same base EG and between the same parallel lines EG and DA.
Therefore,
A(ΔEDG) = A(ΔEGA)
Subtracting ΔEOG from both sides, we have
A(ΔEOD) = A(ΔGOA) ...(i)
Similarly,
A(ΔDPC) = A(ΔBPF) ...(ii)
Now
A(ΔGDF) = A(ΔGOA) + A(ΔBPF) + A(pen. ABPDO)
= A(ΔEOD) + A(ΔDPC) + A(pen. ABPDO)
= A(pen. ABCDE)
Hence, proved.
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