In Fig, $A P ∥ B Q ∥ C R$ . Prove that $a r (A Q C) = a r (P B R)$

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Solution

Given: $A P ∥ B Q$ .
We can say that perpendicular distance from $A$ to $B Q$ is equal to perpendicular distance from $P$ to $B Q$ .
Let it be $d_{1}$ .
$a r (A Q B) = \frac{1}{2} \times B Q \times d_{1}$

$a r (P Q B) = \frac{1}{2} \times B Q \times d_{1}$

$∴$ $a r (A Q B) = a r (P Q B)$

Given: $C R ∥ B Q$
We can say that perpendicular distance from $C$ to $B Q$ is equal to perpendicular distance from $R$ to $B Q$ .
Let it be $d_{2}$
$a r (C Q B) = \frac{1}{2} \times B Q \times d_{2}$

$a r (R Q B) = \frac{1}{2} \times B Q \times d_{2}$

$∴$ $a r (C Q B) = a r (R Q B)$ .

$a r (A Q C) = a r (A Q B) + a r (B Q C) = a r (P Q B) + a r (R Q B) = a r (P B R)$

$∴ a r (A Q C) = a r (P B R)$

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