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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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