AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. 10.26). Show that the line PQ is perpendicular bisector of AB.

Consider the figure,

We have AB is a line segment

P, Q are the points on opposite sides on AB

Such that,

AP = BP (i)

AQ = BQ (ii)

To prove: PQ is perpendicular bisector of AB

Proof: Now, consider ,

AP = BP (From i)

AQ = BQ (From ii)

PQ = PQ (Common)

Therefore, By SSS theorem

(iii)

are isosceles triangles [From (i) and (ii)]

∠PAB = ∠PBA

And,

∠QAB = ∠QBA

Consider,

C is the point of intersection of AB and PQ

PA = PB [From (i)]

∠APC = ∠BPC [From (iii)]

PC = PC (Common)

By SAS theorem,

AC = CB

And, ∠PCA = ∠PCB (By c.p.c.t) (iv)

And also,

ACB is line segment

∠ACP + ∠BCP = 180^o

But, ∠ACP = ∠PCB

∠ACP = ∠PCB = 90^o (v)

We have,

AC = CB

C is the mid-point of AB

From (iv) and (v), we conclude that

PC is the perpendicular bisector of AB

Since, C is the point on line PQ, we can say that PQ is the perpendicular bisector of AB.