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

17:30

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17:13

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13:17

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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= $\frac{A B}{A B}$ = $\frac{A C}{A B}$ + $\frac{B C}{A B}$

Hence, $\frac{B C}{A B}$ = $\frac{13}{30}$ .

$\frac{A C}{B C}$ = $\frac{\frac{A C}{A B}}{\frac{B C}{A B}}$ = $\frac{17}{13}$

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What is the ratio $\frac{A C}{B C}$ for the line segment AB following the construction method below?
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What is the ratio ACBC for the following construction method: A ray is extended from A and 30 arcs of equal lengths are cut, cutting the ray at A1,A2.......A30. A line is drawn from A30 to B and a line parallel to A30B is drawn, passing through the point A17.

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 Hence, BCAB = 1330. ACBC = ACABBCAB = 1713