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...
 A: 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 $x=20/9$.
A: I was waiting for a more elegant solution to come up, but has not, so I will go ahead. Let the triangle be $ABC$, such that $M$ is on $AC$, and $N$ is on $AB$. Let $AM = x$, $AN = y$, and $MN = z$, and $\angle BAC = \theta$. Since, triangles $ABC$ and $ANM$ are similar due to AA criterion, we have the following relations: $y = \frac{4x}{3}$ and $z= \frac{2x}{3}$.   $(1)$
Now, the area of the triangle $AMN$ can be given by $\frac{1}{2}xy\cdot \sin\theta$. Since $BNMC$ is a tangential quadrilateral, its area is given by the formula $rs_1$, where $r$ is the inradius and $s$ is the semiperimeter. Summing both of them gives us the area of triangle $ABC$, which can be given by the formula, $rs_2$, as well, but note that their semiperimeters would be different though the inradius would be the same. We can find that $s_2=9$, which gives us the equation:
$$rs_1 + \frac{1}{2}xy\cdot \sin\theta = rs_2 = 9r$$
So, our first task is to eliminate $\sin\theta$. In order to that, we have to eliminate $s_1$ as well:
$$s_1=\frac{4 + (6 -x)+z+(8-y)}{2} = 9 - \frac{x+y}{2}+\frac{z}{2} =9 - \frac{x+y}{2}+\frac{x}{3}$$     
This gives:
$$ \sin\theta = \frac{2r}{xy}(\frac{x+y}{2} - \frac{x}{3})$$  
Notice that the area of $ABC$ can be given by $\frac{1}{2}\cdot6\cdot8\cdot \sin\theta =24\cdot \sin\theta$. So,
$$24\cdot \sin\theta = \frac{48r}{xy}(\frac{x+y}{2} - \frac{x}{3}) =9r $$  
Cancelling the $r$ [and a bit of manipulation] and substituting for $y$ from $(1)$ gives us:
$$\frac{1}{2}(x+\frac{5x}{3})= \frac{x^2}{4}$$
Solving the [actually linear] equation gives us , $x = \frac{10}{3}$. Now we have reached our final destination, to get the final answer, substitute back from (1), which gives:
$$MN = z = \frac{2x}{3} = \frac{2\cdot\frac{10}{3}}{3}= \frac{20}{9}$$
BTW: Although the length of explanation suggests it took 10 - 20 minutes to solve the question, it took me only 2 - 3. Its surely a long method, but fast and effective nonetheless. When trying to solve questions like this, and not looking to come up with especially elegant solutions, equating [and summing if required] some property of the geometric figures from different aspects will always help, and area is always a good try.
A: The height of ∆ABC = (3√15)/4.2=(3.√15)/3 . Therefore, the height of ∆ CMN = (3.√15)/2-2r = (3.√15)/2-2√15/3 = (5.√15)/6.
So, MN/4=((5√15)/6)/((3.√15)/2)  = 5/(9 ) ,i.e., MN = 20/9.
