Draw a line segment of length $7.6 \, cm$ and divide it in the ratio $5: 8$. Measure the two parts and give the justification of the construction.

(N/A) line segment of length $7.6 \, cm$ can be divided in the ratio of $5: 8$ as follows:
$1.$ Draw a line segment $AB$ of $7.6 \, cm$ and draw a ray $AX$ making an acute angle with line segment $AB$.
$2.$ Locate $13 (= 5 + 8)$ points,$A_1, A_2, A_3, \dots, A_{13}$,on $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = \dots = A_{12}A_{13}$.
$3.$ Join $BA_{13}$.
$4.$ Through the point $A_5$,draw a line parallel to $BA_{13}$ (by making an angle equal to $\angle AA_{13}B$) at $A_5$ intersecting $AB$ at point $C$.
$C$ is the point dividing line segment $AB$ of $7.6 \, cm$ in the required ratio of $5: 8$.
The lengths of $AC$ and $CB$ can be measured. They are $2.9 \, cm$ and $4.7 \, cm$ respectively.
Justification:
By construction,we have $A_5C \parallel A_{13}B$. By applying the Basic Proportionality Theorem $(BPT)$ for the triangle $AA_{13}B$,we obtain:
$\frac{AC}{CB} = \frac{AA_5}{A_5A_{13}}$
From the construction,$AA_5$ contains $5$ equal divisions and $A_5A_{13}$ contains $8$ equal divisions.
Therefore,$\frac{AA_5}{A_5A_{13}} = \frac{5}{8}$.
Thus,$\frac{AC}{CB} = \frac{5}{8}$. This justifies the construction.