Question

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Answer 1

`\text{C.} \ \ \ {\left(\textit{AB}\right)}^(2) \ = \ \left(\textit{AC}\right)\left(\textit{AD}\right)`

Explanation:

This problem uses the concept of the tangent-secant theorem which describes the relationship of the segments a secant line and a tangent line with the associated circle. This theorem is found as Proposition 36 in Book 3 of Euclid's Elements.

As shown in the figure attached below, segment AB (in blue) forms a tangent with the circle BCD and segment AD (in orange) is the secant where it intersects the circle at point C.

Furthermore, let two segments (in green) be drawn one from point C and point D.

To show that
`\triangle ABC` is similar to
`\triangle ADB`, notice that both triangles share a common angle
`\angle BAC`. Additionally, by the alternate segment theorem,
`\angle ABC` is equal to
`\angle ADB`. Therefore,
`\angle ACB` is also equal to
`\angle ABD`.

Hence,
`\triangle ABC` is indeed similar to
`\triangle ADB`. This implies the ratio of the sides of both triangles is the same. Particularly,

`\displaystyle{(AB)/(AD) \ \ = \ \ (AC)/(AB)}`.

Then, performing cross multiplication yields

`{\left(AB\right)}^(2) \ \ = \ \ \left(AC\right)\left(AD\right)`.

Therefore, the product of the lengths of the secant segment and its external segment is equal to the square of the length of the tangent segment.