The point P is equidistant from points A,B. Prove that if M is the midpoint of AB then PM is perpendicular to AB?

1 Answer
Jul 6, 2018

As proved

Explanation:

Given AP = BP and AM = BM To prove hat (AMP) = hat (BMP) = 90^@

"In triangle APB", AP = BP," hence it's an isosceles triangle"

:. hat (PAM) = hat (PBM)

"Since " AM = BM, hat (APM) = hat (BPM), " Equal sides will have equal angle opposite to them theorem"

"We know sum of three angles of a triangle equals " pi^c or 180^@

"Hence, in triangle APM ", pi - (hat (PAM) + hat (APM)) = hat (PMA)

"Similarly in triangle BPM", pi - (hat (PBM) + hat (BPM)) = hat (PMB)

"Since " hat (PAM) + hat (APM) = hat (PBM) + hat (BPM), hat(AMP) = hat (BMP)

But hat (AMP), hat (BMP) " are supplementary angles"

"i.e. " hat (AMP) + hat (BMP) = pi^c " or " 180 ^@

:. hat (AMP) = hat (BMP) = pi/2 " or " 90^@

"That means "bar(PM) " is the perpendicular bisector of "bar(AB)