In $\Delta ABC$,the points $M$ and $N$ are on sides $AB$ and $AC$ respectively such that $AM = \frac{1}{4} AB$ and $AN = \frac{1}{4} AC$. Prove that $MN = \frac{1}{4} BC$.

(N/A) Let $P$ and $Q$ be the midpoints of $AB$ and $AC$ respectively in $\Delta ABC$. Draw $PQ$.
In $\Delta ABC$,$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $AC$. By the Midpoint Theorem,$PQ \parallel BC$ and $PQ = \frac{1}{2} BC$.
Now,$AP = \frac{1}{2} AB$ and $AM = \frac{1}{4} AB$.
Therefore,$AM = \frac{1}{2} AP$.
Similarly,$AQ = \frac{1}{2} AC$ and $AN = \frac{1}{4} AC$.
Therefore,$AN = \frac{1}{2} AQ$.
Thus,in $\Delta APQ$,$M$ is the midpoint of $AP$ and $N$ is the midpoint of $AQ$.
By the Midpoint Theorem,$MN \parallel PQ$ and $MN = \frac{1}{2} PQ$.
Substituting $PQ = \frac{1}{2} BC$ into the equation for $MN$:
$MN = \frac{1}{2} \left( \frac{1}{2} BC \right) = \frac{1}{4} BC$.
Hence,$MN = \frac{1}{4} BC$ is proved.