Question

AB is a line segment and line l is its perpendicular bisector. If P is a point on line l, show that P is equidistant from A and B.

Answer 1

Given:

• AB is a line segment.
• l is the perpendicular bisector of AB.
• P is a point on line l.
• Let Q be the intersection of l and AB so Q lies on AB.

To Prove:

• PA = PB   ...(P is equidistant from A and B)

Proof [Step-wise]:

1. Since l bisects AB at Q.

AQ = QB   ...(Definition of bisector)

2. Because l is perpendicular to AB.

∠AQP = ∠BQP

∠AQP = 90°   ...(Definition of perpendicular)

3. PQ is a common side of triangles ΔPAQ and ΔPBQ.

So PQ = PQ   ...(Common side)

4. In ΔPAQ and ΔPBQ we have:

AQ = BQ   ...(Step 1)

PQ = PQ   ...(Step 3)

And the included angles at Q are equal   ...(Step 2)

5. By SAS congruence

ΔPAQ ≅ ΔPBQ

6. From the congruence

Corresponding sides PA and PB are equal (CPCTC):

PA = PB

So P is equidistant from A and B.