Angle Bisector

In the following example, there is an equilateral triangle $\triangle ABC$.

Note the bisector $BD$ that goes from point $B$ to point $D$ and divides the angle $∡ABC$ into $2$.

That is, angle$∡ABD = 30°$ and angle $∡CBD = 30°$ therefore are equal to each other.

In the following example, a square $ABCD$ is presented.

Note that the bisector $BD$ that goes from point $D$ to point $B$ and divides the angle $∡ADC$ into $2$.

That is, the angle $∡ADB = 45°$ and the angle $∡CDB = 45°$.

In the following example, a rhombus $ACBD$ is presented.

Note that the bisector $CD$ that goes from point $C$ to point $D$ and divides the angle ∡ACB into $2$.

That is, the angle $∡ACD = 45°$ and the angle $∡DCB = 45°$ which are equal.

In this example, a graph is presented with two parallel lines $A$ and $B$.

Note that the bisector $DE$ that goes from point $D$ to point $E$ and divides the angle $∡ADF$ into $2$.

That is, the angle $∡ADE = 25°$ and the angle $∡EDF = 25°$ which are equal.

Example 5: Bisector inside a circle

The line $DB$ intersects with the line $AC$ at point $O$ and forms the angle $∡AOD = 90°$

The bisector $FO$ splits the angle $∡AOD = 90°$ into $2$ equal angles of $45$ degrees.

Having that the angle $∡AOF = 45°$ and the angle $∡FOD = 45°$ are equal.

If you are interested in learning more about other angle topics, you can enter one of the following articles:

• Angle Notation
• Sides, Vertices, and Angles
• Right Angle
• Acute Angle
• Obtuse Angle
• Straight Angle
• Adjacent Angles
• Vertically Opposite Angles
• Alternate Angles
• Corresponding Angles
• Sum of the Angles of a Triangle
• Sum of the Angles of a Polygon
• Sum and Difference of Angles

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In the following example, there is an equilateral triangle $ABC$.

A. Try to draw a new bisector, that divides the angle $∡ABD$ into $2$.

B. Specify the size of the two newly formed angles

Solution to exercise 1:

A. The new bisector $BE$ splits the angle $∡ABD$ into $2$ equal angles of $15°$ degrees each.

B. The size of the formed angles is $∡ABE = 15°$ which is equal to angle $∡EBD = 15°$.

In the following example, a square $ABCD$ is presented.

A. The angle $∡ABC$ is equal to the angle of $∡ADC$. Can it be said that $BD$ serves as the bisector of the angle $∡ABC$?

Solution to exercise 2:

The line $BD$ created $2$ points where the angle was divided into $2$ equal angles.

Therefore, $DB$ is a bisector of the two angles $∡ADC$ and $∡ABC$

Exercise 3: (Bisector within a rhombus )

In the following example, a rhombus $ACBD$ is presented.

Note the bisector $CD$ that goes from point $C$ to point $D$ and divides the angle $∡ACB$ by $2$.

If a bisector is drawn between points $A$ and $B$, and the intersection point between the two lines in the center of the rhombus is $O$.

What type of triangle would triangle $∡AOC$ be?

Solution to exercise 3:

If we draw the bisector from point $A$ to point $B$ when the intersection point between the lines $AB$ and $CD$ will be $O$.

And note that the sum of the angles of a triangle must equal $180°$.

Therefore the triangle $AOC$ will be a right triangle.

This is because:

The angle $∡CAO = 30°$ and the angle $∡OCA = 60°$.

And the sum of the angles of a triangle is equal to $180°$.

Therefore $180° - 90° = 90°$.

And a triangle whose one of its angles is equal to $90°$ is a right triangle.

In this example, a graph is presented with two parallel lines $A$ and $B$.

Note that the bisector $DE$ that goes from point $D$ to point $E$ divides the angle $∡ADF$ by $2$.

That is, in an angle $∡ADE = 25°$ and in an angle $∡EDF = 25°$ which are equal.

In this exercise, we will ask you to draw another line parallel to line $ED$.

Solution to exercise 4

Keep in mind that line $A$ and line $B$ are parallel lines, and are intersected by line $CF$.

Drawing a line $GH$ that is the bisector of angle $∡BKF$.

And to be sure that line $GH$ is parallel to line $ED$, it is enough to see that the angle of $∡GKF$ is equal to $25°$.

If a line is drawn between point $A$ and point $B$, will the new line created $AB$ be parallel to the line $FE$?

Solution to exercise 5:

Given that the two lines $AC$ and $DB$ intersect perpendicularly forming a $90°$ angle between them.

It can be concluded that if we draw a line between the 4 points $ABCD$ we will form a square that is divided in the middle by the line $FE$.

Then the line $FE$ forms a rectangle $ABHG$.

One of the properties of a rectangle is that the opposite sides of the rectangle are parallel to each other.

Therefore, the line $AB$ is parallel to the line $FE$.

What is a bisector?

It is a line segment that passes through the vertex of an angle and divides it into two equal parts.

What is known as a bisector in a triangle?

It is the line segment that divides an interior angle of the triangle into two equal angles.

How many bisectors does a triangle have?

Remember that a triangle has three vertices, therefore, it has three bisectors.

What is the measure of the equal angles generated by the bisectors of an equilateral triangle?

Recalling that the measure of the internal angles of any equilateral triangle is $60°$, then the bisectors will divide these angles into two equal angles of $30°$ each.