How do you prove that a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects?
1 answer:
Check the picture drawn
Consider the line segment AB,
let M be the midpoint of AB, so AM=MA
erect the perpendicular MT to AB from point M. Pick a point P on MT and join it to the points A and B
The triangles PMA and PMB are congruent from the Side Angle Side congruence postulate:
AM=MA, PM is common and m(PMA)=m(PMB)=90°, as MT is perpendicular to AB
so PA=PB
Consider the line segment AB,
let M be the midpoint of AB, so AM=MA
erect the perpendicular MT to AB from point M. Pick a point P on MT and join it to the points A and B
The triangles PMA and PMB are congruent from the Side Angle Side congruence postulate:
AM=MA, PM is common and m(PMA)=m(PMB)=90°, as MT is perpendicular to AB
so PA=PB
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