Question

AB is
is a line segment and line I is its perpendicular bisector. If a point p
lies on I, show that P is equedistant
from A and B​

Answer 1

Answer:

Here is your answer

Step-by-step explanation:

Line l⊥AB and passes through C which is the midpoint of AB

To show that PA=PB

In △PCA and △PCB

AC=BC since C is the mid-point of AB

∠PCA=∠PCB=90∘(given)

PC=PC(common)

So,△PCA≅△PCB by SAS rule.

and so PA=PB as they are corresponding sides of congruent triangles.

HOPE U UNDERSTOOD

MARK ME BRAINLIEST AND FOLLOW ME AND GIVE ME HEART.