Question

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
$\left(i\right)ar\left(△PBQ\right)=ar\left(△ARC\right)\left(ii\right)ar\left(PRQ\right)=\frac{1}{2}ar\left(△ARC\right)\left(iii\right)ar\left(△RQc\right)=\frac{3}{8}ar\left(△ABC\right)$

Answer 1

Construction: Join AQ, PC.

Proof : $\therefore$ CR is the median of $△APC\therefore ar\left(△CRA\right)=ar\left(△CRP\right)=\frac{1}{2}ar\left(△ACP\right)$
Similarly, P is the median of $△ABC△ar\left(△CAP\right)=ar\left(△CPB\right)ar\left(△ACR\right)=\frac{1}{2}ar\left(△CPB\right)$
Again PQ is the median of $△PBC,\therefore ar\left(△CPB\right)=2ar\left(△PBQ\right)$
(ii)
Since QP and QR are the medians of \triangle QAB and \triangle QAP respectively.
$\therefore ar\left(△QAP\right)=ar\left(△QBP\right)andar\left(△QAP\right)=2ar$
From (vi) and (vii \\
ar $\left(△PRQ\right)=\frac{1}{2}ar\left(△PBQ\right)$
(iii) Since Cr is a median of $△CAP\therefore ar\left(△ARC\right)=\frac{1}{2}ar\left(△CAP\right)=\frac{1}{2}\left[\frac{1}{2}ar\right(△ABC\left)\right]=\frac{1}{4}ar\left(△ABC\right)$
Since RQ is a median of $△RBCar\left(△RQC\right)=\frac{1}{2}ar\left(△RBC\right)$
$=\frac{1}{2}\left[ar\right(△ABC)-ar(△ARC\left)\right]=\frac{1}{2}\left[ar\right(triangleABC)-\frac{1}{4}ar(△ABC\left)\right]=\frac{1}{2}\left[\frac{3}{4}ar\right(△ABC\left)\right]=\frac{3}{8}ar\left(△ABC\right)$