Lucia constructed CD¯¯¯¯¯¯¯¯ as a copy of AB¯¯¯¯¯¯¯¯ using a straightedge and compass. What can Lucia say about AB and CD ?
If Lucia constructed CD¯¯¯¯¯¯¯¯ as a copy of AB¯¯¯¯¯¯¯¯ using a straightedge and compass, then she can say that AB and CD are congruent segments.
ucia constructed CD¯¯¯¯¯¯¯¯ as a copy of AB¯¯¯¯¯¯¯¯ using a straightedge and compass. What can Lucia say about AB and CD ?
(1 point)
Responses
AB=2CD.
m Modifying above upper A upper B with bar .
AB=12CD.
m Modifying above upper A upper B with bar .
AB=CD.
m Modifying above upper A upper B with bar .
AB=14CD.
Lucia can say that AB is equal to 14 times CD.
Since Lucia constructed CD¯¯¯¯¯¯¯¯ as a copy of AB¯¯¯¯¯¯¯¯ using a straightedge and compass, this implies that AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ have the same length. In other words, Lucia can say that AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ are congruent.
To analyze what Lucia can say about AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯, let's consider what it means to copy a line segment using a straightedge and compass.
Copying a line segment using a straightedge and compass is a construction technique known as "segment duplication." Here's how it works:
1. Place the compass needle at one endpoint of AB¯¯¯¯¯¯¯¯ and draw an arc that intersects AB¯¯¯¯¯¯¯¯ at some point.
2. Without adjusting the compass, place the compass needle at the other endpoint of AB¯¯¯¯¯¯¯¯ and draw an arc that intersects the first arc at a different point.
3. Use a straightedge to connect the two intersection points, forming CD¯¯¯¯¯¯¯¯.
Now, when Lucia constructed CD¯¯¯¯¯¯¯¯ as a copy of AB¯¯¯¯¯¯¯¯ using this technique, we can make the following observations:
1. Length: CD¯¯¯¯¯¯¯¯ has the same length as AB¯¯¯¯¯¯¯¯. By construction, the compass was set to the length of AB¯¯¯¯¯¯¯¯, so when the arcs were drawn, they must intersect at a distance equal to AB¯¯¯¯¯¯¯¯'s length. Hence, CD¯¯¯¯¯¯¯¯ has the same length as AB¯¯¯¯¯¯¯¯.
2. Collinearity: The three points A, B, and C lie on the same line. Since CD¯¯¯¯¯¯¯¯ is a copy of AB¯¯¯¯¯¯¯¯, it must lie on the same line as AB¯¯¯¯¯¯¯¯. Therefore, the three points A, B, and C are collinear.
In summary, Lucia can confidently say that CD¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯ have the same length, and the points A, B, and C are collinear.