Question

Prove the construction method used for constructing similar triangles of the given ratio of corresponding sides.

Answer 1

$\frac{B{C}^{\prime }}{C^{\prime }C}=\frac{3}{1}$

Therefore, $\frac{BC}{B{C}^{\prime }}=\frac{B{C}^{\prime }+C^{\prime }C}{B{C}^{\prime }}=1+\frac{C^{\prime }C}{B{C}^{\prime }}=1+\frac{1}{3}=\frac{4}{3}$, i.e., $\frac{B{C}^{\prime }}{BC}=\frac{3}{4}$

Also C'A' is parallel to CA. Therefore, $△$A'BC' ~ $△$ ABC.

So, $\frac{A^{\prime }B}{AB}=\frac{A^{\prime }{C}^{\prime }}{AC}=\frac{B{C}^{\prime }}{BC}=\frac{3}{4}$