Construct a triangle with sides $4 \, cm$,$5 \, cm$,and $7 \, cm$. Then,construct a triangle similar to it whose sides have lengths in the ratio $2:3$ to the lengths of the corresponding sides of the first triangle.

(N/A) Data: Construct $\Delta ABC$ in which $AB = 4 \, cm$,$BC = 7 \, cm$,and $AC = 5 \, cm$.
To construct: Construct $\Delta BPQ$ similar to $\Delta ABC$ such that the ratio of their corresponding sides is $2:3$.
Steps of construction:
$(1)$ Construct $\Delta ABC$ in which $AB = 4 \, cm$,$BC = 7 \, cm$,and $AC = 5 \, cm$.
$(2)$ In the half-plane of $\overleftrightarrow{BC}$ that does not contain $A$,draw a ray $\overrightarrow{BX}$ such that $\angle CBX$ is an acute angle.
$(3)$ With an appropriate radius and center $B$,draw an arc to intersect $\overrightarrow{BX}$ at $B_1$. Similarly,with center $B_1$ and the same radius,draw an arc to intersect $\overrightarrow{BX}$ at $B_2$ such that $B-B_1-B_2$. Again,with the same radius and center $B_2$,draw an arc to intersect $\overrightarrow{BX}$ at $B_3$ such that $B_1-B_2-B_3$.
$(4)$ Draw the line segment $\overline{B_3C}$.
$(5)$ Through $B_2$,draw a line parallel to $\overline{B_3C}$ to intersect $\overline{BC}$ at $P$.
$(6)$ Through $P$,draw a line parallel to $\overline{CA}$ to intersect $\overline{AB}$ at $Q$.
Thus,$\Delta BPQ$ is the required triangle.
Justification: $\overleftrightarrow{BX}$ and $\overleftrightarrow{BC}$ are transversals to $\overleftrightarrow{B_3C} \parallel \overleftrightarrow{B_2P}$.
Therefore,$BP:BC = BB_2:BB_3 = 2:3$.
Similarly,$\overleftrightarrow{BC}$ and $\overleftrightarrow{BA}$ are transversals to $\overleftrightarrow{CA} \parallel \overleftrightarrow{PQ}$.
Therefore,$BQ:BA = BP:BC = 2:3$.