Question

AB is a line segment and line lis its perpendicular
bisector. If a point P lies on l, show that P is
equidistant from A and B.
​

Answer 1

Answer:

We have show ∆API congruent to ∆BPI

ok now

IP=IP....... common side

Angle PIA=Angle PIB ....both 90°

AI=BI...........PI is bisector

Therefore ∆API congruent to ∆BPI

now AP=BP......CPCT

Hence proved P is equidastant from A and B

Step-by-step explanation:

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