$ P $ and $ Q $ are respectively the mid-points of sides $ \mathrm{AB} $ and $ \mathrm{BC} $ of a triangle $ \mathrm{ABC} $ and $ \mathrm{K} $ is the mid-point of $ \mathrm{AP} $, show that
(i) $ \operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\mathrm{ARC}) $
(ii) ar $ (\mathrm{RQC})=\frac{3}{8} $ ar $ (\mathrm{ABC}) $
(iii) ar $ (\mathrm{PBQ})=\operatorname{ar}(\mathrm{ARC}) $

Given:

\( P \) and \( Q \) are respectively the mid-points of sides \( \mathrm{AB} \) and \( \mathrm{BC} \) of a triangle \( \mathrm{ABC} \) and \( \mathrm{K} \) is the mid-point of \( \mathrm{AP} \)

To do:

We have to show that

(i) \( \operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\mathrm{ARC}) \)
(ii) ar \( (\mathrm{RQC})=\frac{3}{8} \) ar \( (\mathrm{ABC}) \)
(iii) ar \( (\mathrm{PBQ})=\operatorname{ar}(\mathrm{ARC}) \)

Solution:

We know that,

The median of a triangle divides it into two triangles of equal area.

$PC$ is the median of $\triangle ABC$.

$ar (\triangle BPC) = ar (\triangle APC)$……….(i)

$RC$ is the median of $\triangle APC$.

$ar (\triangle ARC) = \frac{1}{2}ar (\triangle APC)$……….(ii)

$PQ$ is the median of $\triangle BPC$.

This implies,

$ar (\triangle PQC) = \frac{1}{2}ar (\triangle\triangle BPC)$……….(iii)

From (i) and (iii), we get,

$ar (\triangle PQC) = \frac{1}{2} ar (\triangle APC)$……….(iv)

From (ii) and (iv), we get,

$ar (\triangle PQC) = ar (\triangle ARC)$……….(v)

$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.

$PQ \| AC$

$PA = \frac{1}{2}AC$

The triangles between the same parallels are equal in area.

This implies,

$ar (\triangle APQ) = ar (\triangle PQC)$……….(vi)

From (v) and (vi), we get,

$ar (\triangle APQ) = ar (\triangle ARC)$……….(vii)

$R$ is the mid-point of $AP$.

$RQ$ is the median of $\triangle APQ$.

This implies,

$ar (\triangle PRQ) = \frac{1}{2}ar (\triangle APQ)$……….(viii)

From (vii) and (viii), we get,

$ar (\triangle PRQ) = \frac{1}{2}ar (\triangle ARC)$

(ii) $PQ$ is the median of $\triangle BPC$

$ar (\triangle PQC) = \frac{1}{2} ar (\triangle BPC)$

$= \frac{1}{2}\times[\frac{1}{2}ar (\triangle ABC)]$

$= \frac{1}{4}ar (\triangle ABC)$……….(ix)

$ar (\triangle PRC) = \frac{1}{2} ar (\triangle APC)$         [From (iv)]

$ar (\triangle PRC) = \frac{1}{2}\times[\frac{1}{2}ar (\triangle ABC)]$

$= \frac{1}{4}ar (\triangle ABC)$……….(x)

Adding (ix) and (x), we get,

$ar (\triangle PQC) + ar (\triangle PRC) =(\frac{1}{4}+\frac{1}{4})ar (\triangle ABC)$

$ar (PQCR) = \frac{1}{2} ar (\triangle ABC)$……….(xi)

Subtracting $ar(\triangle PRQ)$ from both sides,

$ar (PQCR)-ar (\triangle PRQ) = \frac{1}{2}ar (\triangle ABC)-ar (\triangle PRQ)$

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC) - \frac{1}{2} ar (\triangle ARC)$            (Proved)

$ar (\triangle ARC) = \frac{1}{2} ar (\triangle ABC) -\frac{1}{2} \times [\frac{1}{2}ar (\triangle APC)]$

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC)-(\frac{1}{4})ar (\triangle APC)$

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC)-\frac{1}{4} \times [\frac{1}{2}ar (\triangle ABC)]$        (Since $PC$ is the median of $\triangle ABC$)

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC)-\frac{1}{8}ar (\triangle ABC)$

$ar (\triangle RQC) = [(\frac{1}{2}-(\frac{1}{8})]ar (\triangle ABC)$

$ar (\triangle RQC) = \frac{3}{8}ar (\triangle ABC)$

(iii) $ar (\triangle PRQ) = \frac{1}{2}ar (\triangle ARC)$

$2ar (\triangle PRQ) = ar (\triangle ARC)$……………..(xii)

$ar (\triangle PRQ) = \frac{1}{2}ar (\triangle APQ)$          ($RQ$ is the median of APQ$)……….(xiii)

$ar (\triangle APQ) = ar (\triangle PQC)$

From (xiii) and (xiv), we get,

$ar (\triangle PRQ) = \frac{1}{2} ar (\triangle PQC)$……….(xv)

$ar (\triangle BPQ) = ar (\triangle PQC)$        ($PQ$ is the median of \triangle BPC$)……….(xvi)

From (xv) and (xvi), we get,

$ar (\triangle PRQ) =\frac{1}{2}ar (\triangle BPQ)$……….(xvii)

From (xii) and (xvii), we get,

$2\times(\frac{1}{2})ar(\triangle BPQ)= ar (\triangle ARC)$

$ar (\triangle BPQ) = ar (\triangle ARC)$

Hence Proved.