Theorem 1

Statement

The locus of a point equidistant from two fixed points is the perpendicular bisector of the segment joining the two points.

This theorem is proved in two parts. First prove that any point on the locus satisfies the condition.

Part I

The locus of points equidistant from two given points is the perpendicular bisector of the segment joining them.

Part II

Converse

Any point (or every point) on the perpendicular bisector of the line segment joining two given points is equidistant from them.

Part I

The locus of points equidistant from two given points is the perpendicular bisector of the segment joining them.

Given:

A and B be two points in a plane. P is a moving point in the same plane such that AP = PB.

To prove:

P lies on the perpendicular bisector of AB. (let l be the perpendicular bisector of AB)

Construction:

Join AB. Locate O the mid point of AB.

Join PA and PB.

Join P and O.

Proof:

If P coincides with O then AO = OB or AP = PB.

.

If P does not coincides O, then compare triangles AOP and BOP.

PA=PB ( given)

AO=OB ( O is the mid point of AB)

OP=OP ( common side)

But

( Linear pair)

Part II

Any point (or every point) on the perpendicular bisector of the line segment joining two given points is equidistant from them.

Given:

A and B are two given points in a plane. l is the perpendicular bisector of AB which intersects AB at O.

To prove:

If Q is any point on the perpendicular bisector of AB, AQ=BQ.

Construction:

Join AQ and BQ.

Proof:

If Q coincides with O, then AO = BO.

i.e., AQ=BQ

If Q is distinct i.e., Q does not coincides with O, then

compare triangles AOQ and BOQ.

AO=BO ( O is the mid point of AB)

(given)

OQ=OQ ( common side)

AQ=BQ ( CPCT)