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¯¯¯¯¯¯¯¯
Modifying above upper C upper D with bar perpendicular to Modifying above upper A upper B with bar
1 answer
The correct choice that indicates \( AB \perp CD \) is the perpendicular bisector of \( CD \) if they intersect at \( P \) is:
AB ⊥ CD and CP ≅ DP
This means that \( AB \) is perpendicular to \( CD \), and the segments \( CP \) and \( DP \) are congruent, which defines \( AB \) as the perpendicular bisector of \( CD \).
