Question

What is the ratio $\frac{AC}{BC}$ for the following construction:
A line segment AB is drawn.
A single ray is extended from A and 12 arcs of equal lengths are cut, cutting the ray at $A_1,A_2\dots A_{12}$.
A line is drawn from $A_{12}$ to B and a line parallel to $A_{12}B$ is drawn, passing through the point $A_6$ and cutting AB at C.

• A. 1:2
• B. 1:1
• C. 2:1
• D. 3:1

Answer 1

The correct option is B

1:1

In the construction process given, triangles $△A{A}_{12}B$ and $△A{A}_{6}C$ are similar.

Hence, we get $\frac{AC}{AB}=\frac{6}{12}=\frac{1}{2}$.

By construction $\frac{BC}{AB}=\frac{6}{12}=\frac{1}{2}$.

$\frac{AC}{BC}=\frac{\frac{AC}{AB}}{\frac{BC}{AB}}$

$=\frac{\frac{1}{2}}{\frac{1}{2}}=1.$