Examples of Bisector Example 1: Bisector of an equilateral triangle In the following example, there is an equilateral triangle △ A B C \triangle ABC △ A BC .
Note the bisector B D BD B D that goes from point B B B to point D D D and divides the angle ∡ A B C ∡ABC ∡ A BC into 2 2 2 .
That is, angle∡ A B D = 30 ° ∡ABD = 30° ∡ A B D = 30° and angle ∡ C B D = 30 ° ∡CBD = 30° ∡ CB D = 30° therefore are equal to each other.
Bisector within an equilateral triangle
Example 2: Bisector inside a square In the following example, a square A B C D ABCD A BC D is presented.
Note that the bisector B D BD B D that goes from point D D D to point B B B and divides the angle ∡ A D C ∡ADC ∡ A D C into 2 2 2 .
That is, the angle ∡ A D B = 45 ° ∡ADB = 45° ∡ A D B = 45° and the angle ∡ C D B = 45 ° ∡CDB = 45° ∡ C D B = 45° .
Bisector inside a square
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Question 1 a It is not possible to calculate.
Question 2 a It is not possible to calculate.
Example 3: Bisector within a Rhombus In the following example, a rhombus A C B D ACBD A CB D is presented.
Note that the bisector C D CD C D that goes from point C C C to point D D D and divides the angle ∡ACB into 2 2 2 .
That is, the angle ∡ A C D = 45 ° ∡ACD = 45° ∡ A C D = 45° and the angle ∡ D C B = 45 ° ∡DCB = 45° ∡ D CB = 45° which are equal.
Bisector inside a Rhombus
Example 4: Bisector in a graph with parallel lines In this example, a graph is presented with two parallel lines A A A and B B B .
Note that the bisector D E DE D E that goes from point D D D to point E E E and divides the angle ∡ A D F ∡ADF ∡ A D F into 2 2 2 .
That is, the angle ∡ A D E = 25 ° ∡ADE = 25° ∡ A D E = 25° and the angle ∡ E D F = 25 ° ∡EDF = 25° ∡ E D F = 25° which are equal.
Bisector in a graph with parallel lines
Example 5: Bisector inside a circle Bisector inside a circle
The line D B DB D B intersects with the line A C AC A C at point O O O and forms the angle ∡ A O D = 90 ° ∡AOD = 90° ∡ A O D = 90°
The bisector F O FO FO splits the angle ∡ A O D = 90 ° ∡AOD = 90° ∡ A O D = 90° into 2 2 2 equal angles of 45 45 45 degrees.
Having that the angle ∡ A O F = 45 ° ∡AOF = 45° ∡ A OF = 45° and the angle ∡ F O D = 45 ° ∡FOD = 45° ∡ FO D = 45° are equal.
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Exercises from the previous examples Exercise 1: (Bisector within an equilateral triangle) In the following example, there is an equilateral triangle A B C ABC A BC .
Bisector within an equilateral triangle
A. Try to draw a new bisector, that divides the angle ∡ A B D ∡ABD ∡ A B D into 2 2 2 .
B. Specify the size of the two newly formed angles
Solution to exercise 1:
A. The new bisector B E BE BE splits the angle ∡ A B D ∡ABD ∡ A B D into 2 2 2 equal angles of 15 ° 15° 15° degrees each.
B. The size of the formed angles is ∡ A B E = 15 ° ∡ABE = 15° ∡ A BE = 15° which is equal to angle ∡ E B D = 15 ° ∡EBD = 15° ∡ EB D = 15° .
Exercise 2: (Bisector inside a square) In the following example, a square A B C D ABCD A BC D is presented.
A. The angle ∡ A B C ∡ABC ∡ A BC is equal to the angle of ∡ A D C ∡ADC ∡ A D C . Can it be said that B D BD B D serves as the bisector of the angle ∡ A B C ∡ABC ∡ A BC ?
Bisector inside a square
Solution to exercise 2:
The line B D BD B D created 2 2 2 points where the angle was divided into 2 2 2 equal angles.
Therefore, D B DB D B is a bisector of the two angles ∡ A D C ∡ADC ∡ A D C and ∡ A B C ∡ABC ∡ A BC
Exercise 3: (Bisector within a rhombus ) In the following example, a rhombus A C B D ACBD A CB D is presented.
Note the bisector C D CD C D that goes from point C C C to point D D D and divides the angle ∡ A C B ∡ACB ∡ A CB by 2 2 2 .
If a bisector is drawn between points A A A and B B B , and the intersection point between the two lines in the center of the rhombus is O O O .
What type of triangle would triangle ∡ A O C ∡AOC ∡ A OC be?
Solution to exercise 3:
If we draw the bisector from point A A A to point B B B when the intersection point between the lines A B AB A B and C D CD C D will be O O O .
And note that the sum of the angles of a triangle must equal 180 ° 180° 180° .
Therefore the triangle A O C AOC A OC will be a right triangle .
This is because:
The angle ∡ C A O = 30 ° ∡CAO = 30° ∡ C A O = 30° and the angle ∡ O C A = 60 ° ∡OCA = 60° ∡ OC A = 60° .
And the sum of the angles of a triangle is equal to 180 ° 180° 180° .
Therefore 180 ° − 90 ° = 90 ° 180° - 90° = 90° 180° − 90° = 90° .
And a triangle whose one of its angles is equal to 90 ° 90° 90° is a right triangle.
Exercise 4: (Bisector of an Angle) In this example, a graph is presented with two parallel lines A A A and B B B .
Note that the bisector D E DE D E that goes from point D D D to point E E E divides the angle ∡ A D F ∡ADF ∡ A D F by 2 2 2 .
That is, in an angle ∡ A D E = 25 ° ∡ADE = 25° ∡ A D E = 25° and in an angle ∡ E D F = 25 ° ∡EDF = 25° ∡ E D F = 25° which are equal.
In this exercise, we will ask you to draw another line parallel to line E D ED E D .
Solution to exercise 4
Keep in mind that line A A A and line B B B are parallel lines , and are intersected by line C F CF CF .
Drawing a line G H GH G H that is the bisector of angle ∡ B K F ∡BKF ∡ B K F .
And to be sure that line G H GH G H is parallel to line E D ED E D , it is enough to see that the angle of ∡ G K F ∡GKF ∡ G K F is equal to 25 ° 25° 25° .
Exercise 5
If a line is drawn between point A A A and point B B B , will the new line created A B AB A B be parallel to the line F E FE FE ?
Solution to exercise 5:
Given that the two lines A C AC A C and D B DB D B intersect perpendicularly forming a 90 ° 90° 90° angle between them.
It can be concluded that if we draw a line between the 4 points A B C D ABCD A BC D we will form a square that is divided in the middle by the line F E FE FE .
Then the line F E FE FE forms a rectangle A B H G ABHG A B H G .
One of the properties of a rectangle is that the opposite sides of the rectangle are parallel to each other.
Therefore, the line A B AB A B is parallel to the line F E FE FE .
Do you think you will be able to solve it?
Question 1 a It is not possible to calculate.
Questions on the topic 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 ° 60° 60° , then the bisectors will divide these angles into two equal angles of 30 ° 30° 30° each.
Examples with solutions for Bisector Exercise #1 BD is a bisector.
What is the size of angle ABC?
65 65 65 A A A B B B C C C D D D
Video Solution Step-by-Step Solution Since we are given that the value of angle DBC is 65 degrees, and we know that the angle bisector divides angle ABC into two equal angles, we can calculate the value of angle ABC:
65 + 65 = 130 65+65=130 65 + 65 = 130
Answer Exercise #2 Calculate angle α \alpha α given that it is a bisector.
α α α 60 60 60 A A A a a a
Video Solution Step-by-Step Solution Since an angle bisector divides the angle into two equal angles, and we are given that one angle is equal to 60 degrees. Angle α \alpha α is also equal to 60 degrees
Answer Exercise #3 Which of the following figures has a bisector?
Video Solution Step-by-Step Solution The answer is C because the angle bisector divides the angle into two equal angles. In diagram C, the angle bisector divides the right angle, which is equal to 90 degrees, into 2 angles that are equal to each other. 45 = 45 45=45 45 = 45
Answer Exercise #4 ABCD is a square.
∢ ABC = ? ∢\text{ABC}=\text{?} ∢ ABC = ?
A A A B B B D D D C C C
Video Solution Step-by-Step Solution Due to the fact that all angles in a square are equal to 90 degrees, and BC bisects an angle, we can calculate angle ABC accordingly:
90 : 2 = 45 90:2=45 90 : 2 = 45
Answer Exercise #5 ABCD is a deltoid.
∢ D A C = ? ∢DAC=\text{?} ∢ D A C = ?
A A A B B B C C C D D D 2x 60 2x
Video Solution Step-by-Step Solution As we know that ABCD is a deltoid, and AC is the bisector of an angle and therefore:
B A C = C A D = 2 X BAC=CAD=2X B A C = C A D = 2 X
Now we focus on the triangle BAD and calculate the sum of the angles since we know that the sum of the angles in a triangle is 180 degrees:
2 X + 2 X + 2 X + 60 = 180 2X+2X+2X+60=180 2 X + 2 X + 2 X + 60 = 180
6 X + 60 = 180 6X+60=180 6 X + 60 = 180
180 − 60 = 6 X 180-60=6X 180 − 60 = 6 X
120 = 6 X 120=6X 120 = 6 X
We divide the two sections by 6:120 6 = 6 x 6 \frac{120}{6}=\frac{6x}{6} 6 120 = 6 6 x
20 = x 20=x 20 = x
Now we can calculate the angle DAC:
20 × 2 = 40 20\times2=40 20 × 2 = 40
Answer