Class: 9th | Mathematics (FBISE) | Lecture # | Unit #12 | Theorem #12.1.1 |

Class: 9th |
Mathematics (FBISE) |
Lecture # |
Unit #12 |
Theorem #12.1.1 |
Mathematics Science group |
Any point on the right bisector of a line segment is equidistant from its end points |
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Introduction
In this unit, we will prove theorems and their converses, if
any, about right bisector of a line segment and bisector of an angle.
But before that it will be useful to recall the following definitions:
Right Bisector of a Line Segment
A line is called a right bisector of a line segment if it is perpendicular
to the line segment and passes through its midpoint.
Bisector of an Angle
A ray BP is called the bisector of ∠ABC, if P is a point in the
interior of the angle and m∠ABP = m∠PBC.
Theorem 12.1.1
Any point on the right bisector of a line
segment is equidistant from its end points.
Given
A line LM intersects the line segment AB at
the point C. Such that LM is perpendicular to AB and AC ≅ BC. P is a point on LM.
To Prove
PA ≅ PB
Construction
Join P to the points A and B.
Proof:
In ∆ACP ←→ ∆BCP
AC ≅ BC (given)
∠ACP ≅ ∠BCP (given PC is perpendicular to AB, so that each ∠ at
C = 90^0)

PC ≅ PC (Common
∆ACP ≅ ∆BCP (S.A.S. postulate)
Hence PA ≅ PB (corresponding sides of congruent triangles)
In this unit we stated and will proved the following theorems:
• Any point on the right bisector of a line segment is equidistant
from its end points.
• Any point equidistant from the end points of a line segment is on
the right bisector of it.
• The right bisectors of the sides of a triangle are concurrent.
• Any point on the bisector of an angle is equidistant from its arms.
• Any point inside an angle, equidistant from its arms, is on the bisector of it.
The bisectors of the angles of a triangle are concurrent.
• Right bisection of a line segment means to draw a perpendicular
at the mid point of line segment.
• Bisection of an angle means to draw a ray to divide the given
angle into two equal parts.

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