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?

#### A perpendicular

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

**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

**Solution:**

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).

### Solved Examples:

**1) Find the value of x for the given triangle using the angle bisector theorem.**

**Solution:**

Given that,

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.**

**Solution:**

In the above figure, the line WX is perpendicular bisector to segment ZY.

∴ ∠ WXY= 90° | By perpendicular bisector theorem |

∴ ZX = XY |

Also,

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 | |

= 27 |

Seg WY | = 4x – 5 |

= 4(8) – 5 | |

= 32 – 5 | |

= 27 |

### Unsolved Examples:

**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.**

**Related topics: **

If you have any doubts regarding the article or the examples, please post them in the comments section.

hi, it’s hard to visualise the proofs you’ve put… can u make it in a tabular column. also what is the ‘∴’ symbol mean. plz explain the above.

It’s done. Also, the symbol ‘∴’ is a therefore symbol. Thank you for your feedback.

also for unsolved example #1, how did you get 8?? there is no other number so how do i do it

I have updated it. Thank you