Find length of segment in triangle In triangle $ \bigtriangleup ABC$, the known sides are: $AB=5$, $BC=6$ and $AC=7$. A circle passes through points $A$ and $C$, crosses straight lines $BA$ and $BC$ at points $K$ and $L$, which is non-vertex angle, respectively. Segment KL As to the incircle of the triangle ABC.
Please find the length of the segment $KL$, and show me how to.
 A: 
1. For any $\triangle ABC$ in which 

A circle passes through points $A$ and $C$, crosses straight lines $BA$ and $BC$ at points $K$ and $L$, which is non-vertex angle, respectively

we have that $\triangle ABC \sim \triangle BKL$. Indeed, $AKLC$ is inscribed quadrilateral $\Rightarrow \angle AKL+\angle ACL=\pi$, but also we have $\angle AKL+\angle BKL=\pi \Rightarrow \angle ACL=\angle BKL$; $\angle KBL = \angle ABC$. 
As $\triangle ABC \sim \triangle BKL \Rightarrow |BK|=6x, \ |BL|=5x, |KL|=7x$. So we should find $x$ from additional condition of this problem.
2. I guess, that by 

Segment KL As to the incircle of the triangle ABC

TS means following: For inscribe circle of $\triangle ABC$ - $KL$ is tangent line.

In this case we can write, that $|KL|+|AC|=|AK|+|LC| \ \mathbf{^{*)}}  \Rightarrow$
$\Rightarrow 7x+7=(5-6x) + (6 - 5x)=11-11x \Rightarrow$
$\Rightarrow 18x=4 \Rightarrow x=\frac{2}{9}$.
$|KL|=7x=\frac{14}{9}$
3. $\mathbf{^{*)}}$ As $AKLC$ is tangent quadrilateral thats why |KL|+|AC|=|AK|+|LC|. (See picture).

