A triangle has side lengths 4,6,8. A tangent is drawn to incircle parallel to side 4 cutting ..... Problem : 
A triangle has side lengths $4,6,8$. A tangent is drawn to incircle parallel to side $4$ cutting other two sides at M and N, than length of MN is 
(a) $\frac{10}{9}$
(b) $\frac{20}{ 9}$
(c) $\frac{5}{3}$ 
(d) $\frac{4}{3}$
I have no clue how to proceed such problem however, please help on this I will be greatful to you thanks...

Let the triangle be $ABC$, such that $M$ is on $AC$, and $N$ is on $AB$ and $x=MN$. 
As $ABC$ and $AMN$ are similar we want to derive the scaling factor $k$ by calculating their areas. Let the games begin!
Let $F_1$ be the area of $ABC$, $F_2$ the area of $AMN$ and $F_3$ the area of $NMBC$. From Heron we know $F_1=3\sqrt{15}$ and from $F_1=s\cdot r$ (where $s=(a+b+c)/2$ and $r$ is the radius of the inscribed circle) we get $r=\sqrt{15}/3$.  Now $F_3$ is the area of a trapezoid, so $F_3=\dfrac{x+4}{2}\cdot2r=(x+4)\dfrac{\sqrt{15}}{3}$.  A short computation -- recalling that $F_2=F_1-F_3$ -- reveals that 
$$k^2=\frac{F_2}{F_1}=\frac{5-x}{9}\quad\text{hence}\quad k=\frac{\sqrt{5-x}}{3}.$$
Grand finale: From $x=4k$ we'll happily derive $x=20/9$.
