In this chapter we will understand angle bisector equidistant theorem and learn the method to prove the concept.
What is angle bisector equidistant theorem ?
It says that any point on angle bisector is equidistant from both the arms of the angle.
Consider the angle ∠AOB = 60 degree in above figure.
Line OC is the angle bisector of ∠AOB such that it divides the angle into two equal parts.
Now according to angle bisector equidistant theorem, any point on angle bisector OC is equidistant from the two legs OA and OB.
Here we have selected point P on angle bisector.
According to the theorem, MP = PN
Proof of angle bisector equidistant theorem
Taking the same figure as above;
Consider the triangle OPM and OPN;
∠MOP = ∠NOP = 30 degree { OC is the angle bisector }
∠PMO = ∠PNO = 90 degree { given}
OP = PO { common side }
By AAS congruency condition, triangle OPM and OPN are congruent.
i.e. \mathtt{\triangle OPM\ \cong \triangle OPN}
Since both triangle are congruent we can also say that;
MP = PN
Hence, the point is equidistant from arms OA & OB.
Questions on Angle bisector equidistant theorem
Example 01
In the below figure, OC is the angle bisector of ∠AOB. Find the length of segment MP.
Solution
Given:
OC is the angle bisector.
ON = 8 cm
OP = 10 cm
∠PNO = 90 degree
Applying Pythagoras theorem in triangle PNO.
\mathtt{PO^{2} =\ PN^{2} +ON^{2}}\ \mathtt{10^{2} =PN^{2} +\ 8^{2}}\\\ \\ \mathtt{100-64=\ PN^{2}}\\\ \\ \mathtt{PN^{2} =\ 36}\\\ \\ \mathtt{PN=\ 6\ cm}
P is a point on angle bisector OC and PN is the distance of point P from one of the angle leg OA.
Applying the angle bisector equidistant theorem.
PM = PN = 6 cm
Hence, length of segment PM measures 6 cm.
Example 02
In the below figure OR is the angle bisector of angle AOB. Analyze the below figure and find the value of x.
Solution
Since OP is the angle bisector, we can apply the angle bisector equidistant theorem.
According to the theorem;
MP = PN
x + 3 = 2 (x – 7)
x + 3 = 2x – 14
2x – x = 14 + 3
x = 17
Hence, value of x is 17.