AB is a line segment and line l is its perpendicular bisector. If a po

1.	AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.Achaa sa answer do.....xD
Answer» ________________________________________________ ________________________________________________ $\huge{\underline{\bf{Question:}}}$ AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, SHOW that P is equidistant from A and B. ________________________________________________ ________________________________________________________________________________________________ $\huge{\underline{\bf{Solution:}}}$ since , AB is a line segment, L is drawn perpendicular to AB, A point 'p' lies on line L. (As shown in diagram in the ATTACHMENT) : P is equidistant from A and B. :- In ∆AOP and ∆BOP , OP = OP ( common side ) POA = POB AO = OB ∆AOP ∆BOP (By S.A.S.) => AP = BP ( By C.P.C.T. ) Hence , "P" is equidistant from A and B . $\huge{\underline{\boxed{\bf{Proved!}}}}$ ________________________________________________ ________________________________________________

AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.Achaa sa answer do.....xD

Answer»

$\huge{\underline{\bf{Question:}}}$

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

$\huge{\underline{\bf{Solution:}}}$

since , AB is a line segment,

L is drawn perpendicular to AB,

A point 'p' lies on line L.

(As shown in diagram in the ATTACHMENT)

: P is equidistant from A and B.

In ∆AOP and ∆BOP ,

OP = OP ( common side )

POA = POB

AO = OB

∆AOP ∆BOP (By S.A.S.)

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

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

$\huge{\underline{\boxed{\bf{Proved!}}}}$