Question

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

Answer 1

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

Answer 2

$\textbf{\underline{Question}}$ :-

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

$\textbf{\underline{Solution}}$ :-

since , AB is a line segment,

L is drawn perpendicular to AB,

A point 'p' lies on line L.

( as shown in the diagram in the attachment )

$\bf{To\: prove}$ : P is equidistant from A and B.

$\bf{Prove}$ :-

In ∆AOP and ∆BOP ,

OP = OP ( common side )

$\angle$POA = $\angle$POB

AO = OB

$\therefore$ ∆AOP $\cong$ ∆BOP (By S.A.S.)

=> AP = BP ( By C.P.C.T. )

Hence , "P" is equidistant from A and B .

$\mathcal{\boxed{\red{PROVED}}}$