Consider the figure,
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.