In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$. Prove that $ \operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ARC}) $.
Given:
In a $\triangle ABC, P$ and $Q$ are respectively, the mid-points of $AB$ and $BC$ and $R$ is the mid-point of $AP$.
To do:
We have to prove that \( \operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ARC}) \).
Solution:
Join $AQ$ and $PC$.
$\mathrm{R}$ is mid point of $AP$.
This implies,
$\mathrm{CR}$ is the median of $\triangle \mathrm{APC}$
$\therefore \operatorname{ar}(\Delta \mathrm{CRA})=\operatorname{ar}(\Delta \mathrm{CRP})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ACP})$.........(i)
Similarly,
$CP$ is the median of $\triangle \mathrm{ABC}$
$\therefore \operatorname{ar}(\Delta \mathrm{CAP})=\operatorname{ar}(\Delta \mathrm{CPB})$.......(ii) ($\mathrm{P}$ is the mid point)
From (i) and (ii), we get,
$ar(\Delta \mathrm{ACR})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{CPB})$.......(iii)
$\mathrm{PQ}$ is the median of $\triangle \mathrm{PBC}$,
$\therefore a r(\Delta \mathrm{CPB})=2 \operatorname{ar}(\Delta \mathrm{PBQ})$...........(iv)
From (iii) and (iv), we get,
$\operatorname{ar}(\Delta \mathrm{ARC})=\operatorname{ar}(\Delta \mathrm{PBQ})$......(v)
$\mathrm{QP}$ and $\mathrm{QR}$ are the medians of $\triangle \mathrm{QAB}$ and $\Delta QAP$ respectively.
This implies,
$\operatorname{ar}(\Delta \mathrm{QAP})=\operatorname{ar}(\Delta \mathrm{QBP})$..........(vi)
$\operatorname{ar}(\Delta \mathrm{QAP})=2 a r(\Delta \mathrm{QPR})$......(vii)
From (vi) and (vii)
$\operatorname{ar}(\Delta \mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{PBQ})$
From (v) and (viii), we get,
$\operatorname{ar}(\Delta \mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ARC})$
Hence proved.
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