What is the ratio $\frac{A B}{B C}$ for the line segment AB following the construction method below?
Step 1: A ray is extended from A and 30 arcs of equal lengths are cut, cutting the ray at $A_{1}, A_{2}, . . ., A_{30}$
Step 2: A line is drawn from $A_{30}$ to $B$ and a line parallel to $A_{30} B$ is drawn, passing through the point $A_{17}$ and intersecting AB at C.

17:30

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Solution

The correct option is B
17:13

Here the total number of arcs is equal to m+n in the ratio m:n.

The triangles $△ A A_{17} C$ and $△ A A_{30} B$ are similar.

Hence, $\frac{A C}{A B} = \frac{A A_{17}}{A A_{30}} = \frac{17}{30}$

$1 = \frac{A B}{A B} = \frac{A C}{A B} + \frac{B C}{A B}$
$\frac{B C}{A B} = 1 - \frac{17}{30} = \frac{13}{30}$

Hence, $\frac{A B}{B C} = \frac{17}{13} = 17 : 13$

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The correct option is B 17:13 Here the total number of arcs is equal to m+n in the ratio m:n. The triangles △AA17C and △AA30B are similar. Hence, ACAB=AA17AA30=1730 1=ABAB=ACAB+BCAB BCAB=1−1730=1330 Hence, ABBC=1713=17:13