In $\Delta PQR$,$M$ and $N$ are the midpoints of $PQ$ and $PR$ respectively. $X$ is any point on $QR$. Prove that,$ar(MXN) = \frac{1}{4} ar(PQR)$.

(N/A) $1$. Since $M$ and $N$ are midpoints of $PQ$ and $PR$ respectively,by the Midpoint Theorem,$MN \parallel QR$ and $MN = \frac{1}{2} QR$.
$2$. Consider $\Delta MXN$ and $\Delta M N R$. Both triangles lie between the same parallel lines $MN$ and $QR$.
$3$. The base of $\Delta MXN$ is $MN$ and the base of $\Delta MNR$ is $MN$. Since they share the same base and are between the same parallels,$ar(MXN) = ar(MNR)$.
$4$. In $\Delta PQR$,$MN$ is parallel to $QR$. The height of $\Delta MNR$ with respect to base $MN$ is half the height of $\Delta PQR$ with respect to base $QR$ because $M$ and $N$ are midpoints.
$5$. $ar(MNR) = \frac{1}{2} \times MN \times h_{MNR} = \frac{1}{2} \times (\frac{1}{2} QR) \times (\frac{1}{2} h_{PQR}) = \frac{1}{4} \times (\frac{1}{2} \times QR \times h_{PQR}) = \frac{1}{4} ar(PQR)$.
$6$. Since $ar(MXN) = ar(MNR)$,it follows that $ar(MXN) = \frac{1}{4} ar(PQR)$.