Select the choice that indicates AB←→ is the perpendicular bisector of CD←→ if they intersect at P .(1 point)

Question

Responses

AB¯¯¯¯¯¯¯¯⊥CD¯¯¯¯¯¯¯¯ and CP¯¯¯¯¯¯¯¯≅DP¯¯¯¯¯¯¯¯
Modifying above upper A upper B with bar perpendicular to Modifying above upper C upper D with bar and Modifying above upper C upper P with bar congruent to Modifying above upper D upper P with bar

AC¯¯¯¯¯¯¯¯⊥BD¯¯¯¯¯¯¯¯ and AP¯¯¯¯¯¯¯¯≅BP¯¯¯¯¯¯¯¯
Modifying above upper A upper C with bar perpendicular to Modifying above upper B upper D with bar and Modifying above upper A upper P with bar congruent to Modifying above upper B upper P with bar

AP¯¯¯¯¯¯¯¯≅BP¯¯¯¯¯¯¯¯
Modifying above upper A upper P with bar congruent to Modifying above upper B upper P with bar

CD¯¯¯¯¯¯¯¯⊥AB¯¯¯¯¯¯¯¯

Answer 1

The correct choice that indicates $AB$ is the perpendicular bisector of $CD$ if they intersect at $P$ is:

AB⊥CD and CP≅DP

This means that line segment $AB$ is perpendicular to line segment $CD$, and the lengths of segments $CP$ and $DP$ are equal, confirming that $P$ is the midpoint of $CD$.