Point P lies on the perpendicular bisector of segment AB, meeting AB at M (so PM is perpendicular to AB and AM = MB). Which rule proves triangle PMA congruent to triangle PMB, and hence PA = PB?

Question

Accepted Answer

RHS; PA = PB by CPCT. Right angles at M, AM = MB, PM common: by RHS triangle PMA ≅ triangle PMB, hence PA = PB by CPCT. (SAS also works; here RHS is the right-angle route.)

Point P lies on the perpendicular bisector of segment AB, meeting AB at M (so PM is perpendicular to AB and AM = MB).
Which rule proves triangle PMA congruent to triangle PMB, and hence PA = PB?

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More Class 9 SSS and RHS Congruence practice

Point P lies on the perpendicular bisector of segment AB, meeting AB at M (so PM is perpendicular to AB and AM = MB).Which rule proves triangle PMA congruent to triangle PMB, and hence PA = PB?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 SSS and RHS Congruence practice

Point P lies on the perpendicular bisector of segment AB, meeting AB at M (so PM is perpendicular to AB and AM = MB).
Which rule proves triangle PMA congruent to triangle PMB, and hence PA = PB?