Point on the Bisector of an Angle


Here we will prove that any point on the bisector of an
angle is equidistant from the arms of that angle.

Solution:

Given OZ bisects ∠XOY and PM ⊥ XO and PN ⊥ OY.

Point on the Bisector of an Angle

To prove PM = PN.

Proof:

            Statement

1. In ∆OPM and ∆OPN,

(i) ∠MOP = ∠NOP.

(ii) ∠OMP = ∠ONP = 90°

(iii) OP = OP

2. ∆OPM ≅ ∆OPN.

3. PM = PM. (Proved)

             Reason

1.

(i) OZ bisects ∠XOY.

(ii) Given

(iii) Common side.

2. By AAS criterion.

3. CPCTC.

Note: The abbreviation CPCTC is generally used for
‘Corresponding parts of Congruent Triangles are Congruent’.

From Point on the Bisector of an Angle to HOME PAGE

Didn’t find what you were looking for? Or want to know more information
about
Math Only Math.
Use this Google Search to find what you need.




The math online

Leave a Comment

Your email address will not be published. Required fields are marked *