Locate points on ray $BX$ at equal distances as $B_1,B_2,B_3,\dots$to construct a triangle similar to given $∆\mathrm{ABC}$with $\frac{2}{3}$ sides of that of $∆\mathrm{ABC}$ in such a way that $\angle CBX$is acute. 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 $\frac{2}{3}$ times side $BC.$

$\text{BQ}=\frac{2}{3}\times (\text{BQ}+\text{CQ})$

$\Rightarrow 3\text{BQ}=2\text{BQ}+2\text{CQ}$

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

$\Rightarrow \frac{\text{CQ}}{\text{BQ}}=\frac{1}{2}$

Therefore, Q divides $\mathrm{BC}$ in ratio $2:1.$

So, point $B_3$ should be connected to C, and $B_{2}Q$ should be drawn parallel to$B_{3}C$