A ray $\mathrm{BX}$is drawn such that is $\angle CBX$acute angle and $B_1,B_2,B_3,\dots$are marked at equal distances on BX to construct a triangle similar to given $∆\mathrm{ABC}$with sides equal to $\frac{7}{5}$of the sides of $∆\mathrm{ABC}$ The following points will be joined in the next step:

To divide a line segment in a predetermined ratio.

To divide a line segment in the ratio m: n, where m and n are both positive integers.

$∆\mathrm{PQB}$ is the required triangle.

Since side $BQ$ is $\frac{7}{5}$ times side $BC.$

$\text{BQ}=\frac{7}{5}\times (\text{BQ}+\text{CQ})$

$\Rightarrow 5\text{BQ}=7\text{BQ}+7\text{CQ}$

$\Rightarrow 2\text{BQ}=7\text{CQ}$

$\Rightarrow \frac{\text{CQ}}{\text{BQ}}=\frac{7}{2}$

Therefore, Q divides $\mathrm{BC}$ in ratio $7:2.$

$B_7,Q$ are joined and then $B_5,C$