In the following, write True or False and explain your answer: To make a triangle that is similar to a given $∆\mathrm{ABC}$with sides$\frac{7}{3}$of the corresponding sides of $∆\mathrm{ABC}$draw a $BX$ray that makes an acute angle with $BC$and X lies on the opposite side of $A$with respect to $BC.$ The points $B_1,B_2,B_3,\dots .B_7$are located at equal distances on $BX,B_3$is joined to $C,$and a line segment $B6,C^{\text{'}}$ is drawn parallel to ${\mathrm{B}}_3\mathrm{C}$where ${\mathrm{C}}^{\text{'}}$lies on $\mathrm{BC}$produces. Finally, a line segment ${\mathrm{A}}^{\text{'}}{\mathrm{C}}^{\text{'}}$is drawn parallel to $\mathrm{AC}$

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 $(7>3)$.

Construction: (i) Draw given triangle with given specifications.

(ii) Draw an acute angle CBX.

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

(iv) Produce $\mathrm{BC}$ and $\mathrm{BA}$ as shown id figure.

(v) Join $B_{3}C$.

(vi) Draw $B_7{C}^{\text{'}}\Vert B_{3}C$. is on $BC$produced.

(vii) Draw $C\text{'}A\Vert A{C}^{\text{'}}{A}^{\text{'}}$  on $BA$ produced $∆A^{\text{'}}{B}^{\text{'}}$ is required triangle i.e., $\frac{∆{\mathrm{A}}^{\text{'}}\mathrm{BC}}{\Delta \mathrm{ABC}}=\frac{3}{7}$

Here, ${\mathrm{B}}_7{\mathrm{C}}^{\text{'}}\Vert {\mathrm{B}}_3\mathrm{C}$. But in question

${\mathrm{B}}_6{\mathrm{C}}^{\text{'}}\Vert {\mathrm{B}}_3\mathrm{C}$, which is false.