Question:
To construct a triangle similar to a given $\triangle \mathrm{ABC}$ with its sides $\frac{3}{7}$ of the corresponding sides of $\triangle \mathrm{ABC}$,
first draw a ray $B X$ such that $\angle C B X$ is an acute angle and $X$ lies on the opposite side of $A$ with respect to $B C$. Then, locate points
$B_{1}, B_{2}, B_{3} \ldots$ on $B X$ at equal distances and next step is to join
(a) B 10 to C
(b) $B_{13}$ to $C$
(c) $B_{7}$ to $C$
(d) $B_{4}$ to $C$
Solution:
(c) Here, we locate points B1, B2, B3, B4, B5, B6 and B7 on BX at equal distance and in next step join the last points is B7 to C.