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.


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Answer 1

ANSWER

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,

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

Hence proved.

hope it helps....

Answer 2

$\huge\underline\mathfrak\orange{Question}$

Line l⊥AB and passes through C which is the midpoint of AB. To show that PA=PB.

$\huge\underline\mathcal\red{Answer}$

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.

$\huge\underline\mathcal\purple{BTS \: EXO}$