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Triangles on the Same Base and between the Same Parallels

The side AB o...

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Question 9
The side AB of a parallelogram ABCD is produced to any point P.
A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. (see the following figure).
Show that ar(ABCD) = ar(PBQR).

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Solution

Let us join AC and PQ.
$Δ A C Q$ and $Δ A Q P$ are on the same base AQ and between the same parallels AQ and CP.
$∴ a r (Δ A C Q) = a r (Δ A P Q)$

$a r (Δ A C Q) - a r (Δ A B Q) = a r (Δ A P Q) - a r (Δ A B Q)$
[Subtracting ar $(Δ A B Q)$ from both sides]
$\Rightarrow a r (Δ A B C) = a r (Δ Q B P) . . . . . . . (1)$

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,
$a r (Δ A B C) = \frac{1}{2} a r (A B C D) . . . (2)$

$a r (Δ Q B P) = \frac{1}{2} a r (P B Q R) . . . (3)$

From equations (1), (2), and (3), we obtain;
$\frac{1}{2} a r (A B C D) = \frac{1}{2} a r (P B Q R)$

$\Rightarrow a r (A B C D) = a r (P B Q R)$

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Triangles on the Same Base and between the Same Parallels

Standard IX Mathematics

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