Locate points ${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3,\dots .$ to construct a triangle similar to given $\Delta ABC$with sides $\frac{4}{5}$of that of $\Delta ABC$ on ray $BX$at equal distances, so that $\angle \mathrm{ABX}$is acute. Find the following points that 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 $\frac{4}{5}$times side BA.

$\text{BQ}=\frac{4}{5}\times (\text{BQ}+\text{AQ})$

$\Rightarrow 5\text{BQ}=4\text{BQ}+4\text{AQ}$

$\Rightarrow \text{BQ}=4\text{AQ}$

$\Rightarrow \frac{\text{AQ}}{\text{BQ}}=\frac{1}{4}$

Therefore, Q divides BA in ratio 4: 1 .

So, point ${\mathrm{X}}_5$ should be connected to $\mathrm{A}$, and ${\mathrm{X}}_4\mathrm{Q}$should be drawn parallel to${\mathrm{X}}_5 \mathrm{A}$.