Unseen Theorem Based on triangle In $\triangle ABC$, $P$ & $Q$ are the mid points of $AB$ and $AC$, $S$ is the mid point of $PQ$ and $R$ is any point on $BC$, prove that :$8\triangle SQR=\triangle ABC$.

When I joined $P,C$ then I got $\triangle APC=\triangle BPC=\frac {1}{2} \triangle ABC$
But I couldn't move further from here. So please help me. 
 A: Start with similarity: $\triangle ABC \sim \triangle APQ$ with dimension ratio $2$, and therefore area ratio $2^2 = 4$.
Join $P$ to $R$ and observe $\triangle PSR = \triangle SQR$ (same base, same height "SBSH") If you let $\triangle SQR = x$, then $\triangle PRQ = 2x$.
Now $\triangle APQ = \triangle PRQ$ (SBSH, since the base $PQ$ is common and the height is the same because of the height of $\triangle APQ$ is half the height of $\triangle ABC$ as observed above).
Hence $\triangle APQ = 2x$ and $\triangle ABC = 4(2x) = 8x$ so $\triangle ABC = 8\triangle SQR \ \ (QED)$
A: 
Let X be the mid point of BC.
Join PX and QX.
Now,P,Q,X are the midpoints of the sides of the triangle PQX.
Therefore $ar(\triangle APQ)$=$ar(\triangle PXQ)$=$ar(\triangle PBX)$=$ar(\triangle QXC)$.(by mid-point theorem).
Let,area of each triangle mentioned above =$x$ sq. units.
So,total area of triangle=$x+x+x+x=4x$ sq.units
Also,note that $ar(\triangle SRQ)$=$ar(\triangle SXQ)$.(because they lie on same base and between same parallels).
$SX$is the median of $ar(\triangle PXQ)$.So, $ar(\triangle SXQ)=\frac {x}{2}$sq.units=$ar(\triangle SRQ)$.
So,$$\frac{ar(\triangle ABC)}{ar(\triangle SRQ)}=\frac{4x}{x/2}=\frac{8}{1}$$ .
Hence proved.
