Question

In a $△ABC$,$E$ and $F$ are the mid-point $AB$ and $AC$ respectively. Point $M$ and $N$ 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$

Answer 1

Now, from $△ABC$ we have, $E$ & $F$ one mid-points of $AB$ & $AC$ respectively.

Then we have, $EF\parallel BC$ & $EF=1/2BC----\left(1\right)$

Again from $△AEF$ we have,

$M$ & $N$ are mid points of $AE$ & $AF$ Then we get, $MN\parallel EF$ & $MN=1/2EF----\left(2\right)$

Using $\left(1\right)$ in $\left(2\right)$ we get, $MN=1/2\times \frac{1}{2}\times BC$

$=\left(\frac{1}{4}BC\right)$