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Triangles on the Same Base and between the Same Parallels

ABCD is a tra...

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Question 13
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

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Solution

It can be observed that $Δ A D X$ and $Δ A C X$ lie on the same base AX and are between the same parallels AB and DC.
$Δ A r e a (Δ A D X) = A r e a (Δ A C X) . . . (1)$
$Δ A C Y$ and $Δ A C X$ lie on the same base AC and are between the same parallels AC and XY.
$∴ A r e a (Δ A C Y) = A r e a (A C X) . . . (2)$
From equations (1) and (2), we obtain, $A r e a (Δ A D X) = A r e a (Δ A C Y)$

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Triangles on the Same Base and between the Same Parallels

Standard IX Mathematics

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