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Question
In quadrilateral ABCD given below, AC is the diagonal and 𝐃𝐄 ∥ 𝐀𝐂. Also, 𝐃𝐄 meets 𝐁𝐂 produced at 𝐄. Show that 𝐚𝐫(𝐀𝐁𝐂𝐃) = 𝐚𝐫(△𝐀𝐁𝐄).
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Solution
Given: ABCD is a quadrilateral, AC is diagonal And DE ∣∣ AC and also DE meats BC produced at E.
As per figure ar(ABCD) = ar(ΔABC) + ar(ΔDAC).......................(1)
ΔDAC and ΔEAC lie on same base and between the parallel line DE and AC
Then base and height of both triangle are equal.
ar(ΔDAC) = ar(ΔEAC)..................................(2)
Add ar(ΔABC) to both side of the equation (2), we get
∴ar(ΔDAC) + ar(ΔABC) = ar(ΔEAC) + ar(ΔABC)
Hence ar(ABCD) = ar(ΔABE) (From (2))
As per figure ar(ABCD) = ar(ΔABC) + ar(ΔDAC).......................(1)
ΔDAC and ΔEAC lie on same base and between the parallel line DE and AC
Then base and height of both triangle are equal.
ar(ΔDAC) = ar(ΔEAC)..................................(2)
Add ar(ΔABC) to both side of the equation (2), we get
∴ar(ΔDAC) + ar(ΔABC) = ar(ΔEAC) + ar(ΔABC)
Hence ar(ABCD) = ar(ΔABE) (From (2))
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