Question

In the adjoining figure, 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. Show that the line PQ is the perpendicular bisector of AB.

Answer 1

Given:

• AB is a line segment.
• P and Q lie on opposite sides of AB.
• PA = PB and QA = QB.

To Prove:

• The line PQ is the perpendicular bisector of AB (i.e. PQ ⟂ AB and PQ meets AB at its midpoint).

Proof (Step-wise):

1. Construct MH: Let M be the foot of the perpendicular from P to AB.

So PM ⟂ AB and M lies on AB.

2. Consider right triangles PAM and PBM.

PA = PB   ...(Given)

PM = PM   ...(Common)

∠PMA = ∠PMB = 90°   ...(By construction)

Therefore, triangles PAM and PBM are congruent by RHS (right-angle, hypotenuse, side).

Hence AM = MB.

So M is the midpoint of AB and PM is a perpendicular bisector of AB.

3. Similarly, construct N as the foot of the perpendicular from Q to AB.

So QN ⟂ AB and N lies on AB.

By the same argument RQ: QA = QB, QN common, right angles, triangles QAN and QBN are congruent, so AN = NB. 

Thus, N is the midpoint of AB and QN is a perpendicular bisector of AB.

4. The midpoint of a segment is unique, hence M = N.

Therefore, the two perpendicular bisectors PM and QN coincide; both P and Q lie on the same line through the midpoint of AB, which is perpendicular to AB.

5. Therefore, the line through P and Q (i.e. PQ) is that perpendicular bisector: PQ passes through the midpoint of AB and is perpendicular to AB.

PQ is the perpendicular bisector of AB. It meets AB at its midpoint and is perpendicular to AB.