$P$ and $Q$ are respectively the mid-points of sides $A B$ and $B C$ of a triangle $A B C$ and $R$ is the mid-point of $A P$ , show that:
$a r (R Q C) = \frac{3}{8} a r (A B C)$

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Solution

Median and traingle concept :
As, $C R$ is a median of $∆ C A P$
$a r (A R C) = \frac{1}{2} a r (C A P) = \frac{1}{2} [\frac{1}{2} a r (A B C)] = \frac{1}{4} a r (A B C)$ .
Since, $R Q$ is a median of $∆ R B C$
$a r (R Q C) = \frac{1}{2} a r (R B C) = \frac{1}{2} [a r (A B C) - a r (A R C)] = \frac{1}{2} [a r (A B C) - \frac{1}{4} a r (A B C)] = \frac{1}{2} [\frac{3}{4} a r (A B C)] = \frac{3}{8} a r (A B C)$
Hence, it is proved that $ar (RQC) = \frac{3}{8} ar (ABC)$ .

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