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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.

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Consider the figure,

We have

AB is a line segment and P,Q are points on opposite sides of AB such that

We have to prove that PQ is perpendicular bisector of AB.

Now consider ΔPAQand ΔPBQ,

Now, we can observe that ΔAPB and ΔABQ are isosceles triangles. (From 1 and 2)

⇒ ∠PAB = ∠PBA and ∠QAB=∠QBA

Now consider ΔPAC and  PB,

C is the point of intersection of AB and PQ.

So, from SAS congruency of triangle ΔPAC≅ ΔPBC 

⇒ AC = CB and ∠PCA =∠PCB ......(4)

[ ∵ Corresponding parts of congruent triangles are equal]

And also, ACB is line segment

We have AC = CB ⇒ C is the midpoint of AB

From (4) and (5)

We can conclude that PC is the perpendicular bisector of AB

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

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