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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.
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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