In this post, we will explore fundamental geometry concepts such as the perpendicular bisector and angle bisector. These are the following topics we will study in this article:
- What is a perpendicular bisector?
- The perpendicular bisector theorem
- What is an angle bisector?
- The angle bisector theorem
- Examples of both angle bisector and perpendicular bisector
What is a perpendicular bisector?
Let us gradually break down the perpendicular bisector by first defining what perpendicular is. If two distinct lines, rays, or line segments intersect at 90° or form a right angle with each other, it is called perpendicular lines.
The above figure shows that the line segment AB intersects the line segment CD at point F, thus forming a right angle. Hence, they’re called perpendicular lines.
A bisector is an object (line, line segment, or ray) that intersects another object or line segment in such a way that the segment is divided into two equal parts. Also, a bisector cannot bisect a Line as Line does not have a finite length.
Let’s take a look at the example above to understand how a bisector works. In the above figure, seg AB bisects seg CD such that it divides the segment into two equal parts.
A perpendicular bisector
Once we understand what a perpendicular line and bisector are, defining a perpendicular bisector becomes simple.
A perpendicular bisector is a line, line segment or ray that bisects a segment at a right angle and divides the segment into two equal parts. In short, a perpendicular bisector is a combination of a perpendicular line and a bisector.
Further to know how to construct a perpendicular bisector, I recommend you watch this following video 👇
Perpendicular bisector theorem
Furthermore, by combining all these points, we may finally comprehend the perpendicular bisector theorem.
Statement: Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
Given: line l is the perpendicular bisector of seg AB at point M. Point P is any point on l.
To prove: PA = PB
Construction: Draw Seg AP and Seg BP
|Proof:||In ∆ PMA and ∆ PMB|
|Seg PM ≅ Seg PM||Common side|
|∠PMA ≅ ∠PMB||Each is a right angle|
|Seg AM ≅ Seg BM||Given angle|
|∴ ∆ PMA ≅ ∆ PMB||SAS test (side angle side test)|
|∴ Seg PA ≅ Seg PB|
|∴ l (PA) = l (PB)|
Hence every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
Converse of perpendicular bisector theorem
Statement: Any point equidistant from the end points of a segment lies on the perpendicular bisector of the segment.
Given: Point P is any point equidistant from the end points of seg AB. That is, PA = PB.
To prove: Point P is on the perpendicular bisector of seg AB.
Construction: Take mid-point M of seg AB and draw line PM.
|Proof:||In ∆ PAM and ∆ PBM|
|seg PA ≅ seg PB||perpendicular bisector theorem|
|seg AM ≅ seg BM||midpoint|
|seg PM ≅ seg PM||common side|
|∴ ∆ PAM ≅ ∆ PBM|
|∴ ∠ PMA ≅ ∠ PMB|
|But ∠ PMA + ∠PMB = 180°|
|∠ PMA + ∠PMA = 180°|
|2∠ PMA = 180°|
|∴ ∠ PMA = 90°|
|∴ seg PM ⊥ seg AB|
|But Point M is the midpoint of seg AB||according to construction|
Therefore, line PM is the perpendicular bisector of seg AB. So, point P is on the perpendicular bisector of seg AB.
What is an Angle bisector?
Just like how a bisector divides a line segment into two equal halves, an angle bisector is a ray, line, or line segment that divides an angle into two equal parts.
Construction of an angle bisector
Please refer to the below video to visualize the construction of an angle bisector.
Angle bisector theorem
Statement: If a point is on the angle bisector, then it is equidistance from the sides of the angle.
|Given:||Ray A is the bisector of ∠BAC|
|Point D is any point on Ray A.|
To find: Seg ED ≅ Seg DF
We know that Ray A bisects ∠BAC
|∴ Ray AB ⊥ Seg ED and Ray AC ⊥ Seg DF|
According to the figure,
|∠ AED and ∠ AFD are right angles||all perpendicular lines are right angles|
|∴ ∠ AED ≅ ∠ AFD||all right angles are congruent|
|∴ ∠ BAD ≅ ∠ FAD||definition of angle bisector|
|∴ AD ≅ AD||common side|
|∆ AED ≅ ∆ AFD||AAS|
|∴ Seg ED ≅ Seg DF||corresponding parts of congruent triangle is also congruent|
Hence, it is proved that if a point on the angle bisector (e.g., D), then it is equidistant from the side of the angles (seg ED ≅ seg DF).
1) Find the value of x for the given triangle using the angle bisector theorem.
|PD = 12||PQ = 18||QR = 24||DR = x|
According to angle bisector theorem,
|PD/PQ = DR/QR|
Now substitute the values, we get
|12/18 = x/24|
|X = (⅔)24|
|x = 2(8)|
|x = 16|
Hence, the value of x is 16.
2) Find x and length of each segment.
In the above figure, the line WX is perpendicular bisector to segment ZY.
|∴ ∠ WXY= 90°||By perpendicular bisector theorem|
|∴ ZX = XY|
|Seg WZ ≅ Seg WY||By perpendicular bisector theorem|
|∴ 2x + 11 = 4x – 5||Given|
|∴ 16 = 2x|
|∴ X = 16/2|
|∴ X = 8|
Length of segments
|Seg WZ||= 2x + 11|
|= 2(8) + 11|
|= 16 + 11|
|Seg WY||= 4x – 5|
|= 4(8) – 5|
|= 32 – 5|
1) Find the value of x in ∆ ABC.
2) In ∆ ABC pictured below, AD is the angle bisector of ∠ A. If CD = 9, CA = 12 and AB = 16, find BD.
If you have any doubts regarding the article or the examples, please post them in the comments section.