Question

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

Answer 1

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)`