To construct a triangle similar to a given $∆\mathrm{ABC}$ with its sides in a ratio of $\frac{7}{3}$ with the corresponding sides of $∆ABC$. To draw a ray $BX$that makes an acute angle with$BC$and X lies on the opposite side of $A$ with respect to $BC,$ The points ${\mathrm{B}}_1,{ \mathrm{B}}_2,\dots ,{\mathrm{B}}_7$ are located at equal distances on $\mathrm{BX},{\mathrm{B}}_3$ is joined to C

To build a triangle similar to a given triangle using the given scale factor.

This structure involves two distinct situations.

In one, the triangle to be built is smaller than the given triangle, while in the other, it is larger.

The scale factor denotes the ratio of the sides of the triangle to be constructed to the sides of the given triangle.

Given ratio is $\frac{7}{3}>1$ so, the resulting triangle will be larger than given as ${\mathrm{B}}_7\mathrm{C}\Vert {\mathrm{B}}_3\mathrm{C}$ and $\mathrm{BX}$ is equally divided into 7 part as $(7>3)$.

Steps of Construction:

Draw the given triangle with the given specifications.

Draw an acute angle $\angle CBX$.

Divide $BX$ into 7 equal parts and mark them $B_1,B_2,B_3,\dots ,B_7$.

Produce $\mathrm{BC}$ and $\mathrm{BA}$ as shown in the figure.

Join $B_{3}C$.

Draw $B_7{C}^{\text{'}}\Vert B_{3}C$. to intersect on $BC$ produced.

Draw ${\mathrm{C}}^{\text{'}}\mathrm{A}\Vert \mathrm{AC}·{\mathrm{A}}^{\text{'}}$ on $\mathrm{BA}$ produced $∆{\mathrm{A}}^{\text{'}}{\mathrm{BC}}^{\text{'}}$ is required triangle.

Hence, the given statement is true.