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# In figure below, $AP \| BQ \| \mathrm{CR}$. Prove that $ \operatorname{ar}(\mathrm{AQC})=\operatorname{ar}(\mathrm{PBR}) . $

"

Given:

$AP \| BQ \| \mathrm{CR}$.

To do:

We have to prove that \( \operatorname{ar}(\mathrm{AQC})=\operatorname{ar}(\mathrm{PBR}) . \)

Solution:

$\triangle BQC$ and $\triangle BQR$ lie on the same base $BQ$ and between the parallels $BQ$ and $CR$.

Therefore,

$ar(\triangle BQR) = ar(\triangle BQC)$.....…(i)

$\triangle AQB$ and $\triangle PBQ$ lie on the same base $BQ$ and between the parallels $BQ$ and $AP$.

Therefore,

$ar(\triangle ABQ) = ar(\triangle PBQ)$.....…(ii)

From the given figure, we get,

$ar (\triangle PBR) = ar (\triangle PBQ) + ar (\triangle QBR)$......…..(iii)

$ar(\triangle AQC) = ar (\triangle AQB)+ ar (\triangle BQC)$.......…(iv)

Adding (i) and (ii), we get,

$ar (\triangle BQC + ar(\triangle AQB) = ar (\triangle QBR) + ar (\triangle PBQ)$

$ar(\triangle AQC) = ar(\triangle PBR)$ [From (iii) and (iv)]

Hence proved.

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