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Figures on the Same Base and Between the Same Parallels

In the figure...

Question

In the figure, $A B C D$ is a quadrilateral. $A C$ is the diagonal and $D E | | A C$ and also $D E$ meets $B C$ produced at $E$ . Show that $a r (A B C D) = a r (△ A B E)$ .

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Solution

Given: $A B C D$ is a quadrilateral, $A C$ is diogonal And $D E | | A C$ and also $D E$ meats $B C$ produced at $E .$

As per figure $a r e a (A B C D) = a r e a (Δ A B C) + a r e a (Δ D A C)$ .......................(1)

$Δ D A C$ and $Δ E A C$ lie on same base and between the parallel line DE and AC

Then base and height of both triangle are equal

$a r e a (Δ D A C) = a r e a (Δ E A C)$ ..................................(2)

Add $a r e a (Δ A B C)$ in (2) both side we get

$∴ a r e a (Δ D A C) + a r e a (Δ A B C) = a r e a (Δ E A C) + a r e a (Δ A B C)$

Hence $a r e a (A B C D) = a r e a (Δ A B E)$ (From (2))

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