To construct a triangle similar to a given $\triangle ABC$ with its sides $\frac{3}{7}$ of the corresponding sides of $\triangle ABC$,first draw a ray $BX$ such that $\angle CBX$ is an acute angle and $X$ lies on the opposite side of $A$ with respect to $BC$. Then locate points $B_{1}, B_{2}, B_{3}, \dots$ on $BX$ at equal distances and the next step is to join
- A$B_{10}$ to $C$
- B$B_{7}$ to $C$
- C$B_{13}$ to $C$
- D$B_{4}$ to $C$
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