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Constructing, copying and bisecting angles
Constructing an angle bisector
Example 1
Construct the bisector of angle \latex{\alpha}.
Solution
During the construction, the intersections of points \latex{ A } and \latex{ B } on the arms of the angle and the arc drawn around the vertex of the angle, together with vertex \latex{ O }, form an isosceles triangle. The perpendicular bisector of the base also bisects angle \latex{ α }. Therefore, it is sufficient to construct the perpendicular bisector of line segment \latex{ AB }.
Sketch:
The steps of construction:
- Construct angle \latex{\alpha}. Draw an arc with an arbitrary radius around the centre of the angle's vertex to mark points \latex{ A } and \latex{ B } on the arms of the angle.
- From points \latex{ A } and \latex{ B }, draw intersecting arcs with the same radius as that of the arc constructed in the previous step. Mark the intersection of the arcs with the letter \latex{ M }.
- Connect vertex \latex{ O } with point \latex{ M }. The resulting ray \latex{ f } is the angle bisector of angle \latex{\alpha}.
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ A }
\latex{ 0 }
\latex{ B }
\latex{ A }
\latex{ 0 }
\latex{ B }
\latex{ B }
\latex{ 0 }
\latex{ A }
\latex{ M }
\latex{ 0 }
\latex{ A }
\latex{ M }
\latex{ B }
\latex{ f }
\latex{\alpha}
Note:
It is not necessary to construct line segment \latex{ AB }, as it is sufficient to mark points \latex{ A } and \latex{ B } on the arms of the angle.
Copying angles
Example 2
Angle \latex{\alpha} in the image is a convex angle. Construct angle \latex{\alpha } on a ray \latex{ f } with starting point \latex{ F }.
\latex{ \alpha }
Solution
Mark a point \latex{ A } on one of the arms and a point \latex{ B } on the other. The two points must be at the same distance from vertex \latex{ O }. The resulting \latex{ OAB } triangle is an isosceles triangle. Reconstruct this triangle using its three sides in such a way that point \latex{ O } should be located at point \latex{ F } of ray \latex{ f }, and arm \latex{ OA } should be on ray \latex{ f }. Triangles \latex{ FCD } and \latex{ OAB } are congruent; therefore, the angle at vertex \latex{ F } is \latex{ α }.
Sketch:
Concave angles
Convex angles
The steps of construction:
- Construct angle \latex{\alpha} and ray \latex{ f } with starting point \latex{ F }.
- Draw arcs with the same radii around vertex \latex{ O } of angle \latex{\alpha} and point \latex{ F } of ray \latex{ f }. The intersections of the arms of angle \latex{\alpha} and the arc should be points \latex{ A } and \latex{ B }, while the intersection of the arc and ray \latex{ f } should be point \latex{ C }.
- Measure the length of line segment \latex{ AB } with your compass and intersect the arc from point \latex{ C } using the distance measured with the compass (point \latex{ D }).
- Draw ray \latex{ FD }, which will be the other arm of the angle. \latex{DFC \measuredangle = \alpha}.
\latex{ 0 }
\latex{\alpha}
\latex{ F }
\latex{ f }
\latex{ 1 }
\latex{ 2 }
\latex{ 4 }
\latex{ 3 }
\latex{ 0 }
\latex{ B }
\latex{\alpha}
\latex{ A }
\latex{ C }
\latex{ F }
\latex{ f }
\latex{ 0 }
\latex{\alpha}
\latex{ B }
\latex{ A }
\latex{ D }
\latex{\alpha}
\latex{ C }
\latex{ f }
\latex{ f }
\latex{ C }
\latex{ F }
\latex{ F }
\latex{ 0 }
\latex{\alpha}
\latex{ B }
\latex{ A }
Note:
In the solution above, an arbitrary convex angle was copied.
Copying a concave angle can be done similarly, as a convex and a concave angle together form a full angle.
\latex{\alpha}
\latex{\beta}
\latex{ A }
\latex{ B }
\latex{ \alpha \gt 180° }
Example 3
You have two angles: angle \latex{\alpha} and angle \latex{\beta (\alpha \gt \beta)}.
Construct
- the sum of the two angles \latex{(\alpha + \beta)};
- the difference of the two angles \latex{(\alpha - \beta)}.
\latex{\alpha}
\latex{\beta}
Solution
The construction is similar to copying angles.
The steps of construction:
- Copy angle \latex{ β } next to angle \latex{ α } so that one of the arms of angle \latex{ β } overlaps one of the arms of angle \latex{ α }, while the other arm of angle \latex{ β } is located outside angle \latex{ α }.
- Copy angle \latex{ β } next to angle \latex{ α } so that one of the arms of angle \latex{ β } overlaps one of the arms of angle \latex{ α }, while the other arm of angle \latex{ β } is located inside angle \latex{ α }.
Sketch:
a)
\latex{\alpha}
\latex{\alpha+\beta}
\latex{\beta}
b)
\latex{\alpha}
\latex{\beta}
\latex{\alpha-\beta}
Note:
The method above can be used to double, triple, etc. any arbitrary angle by copying the given angle next to itself several times.
\latex{2\alpha}
\latex{\alpha}
Constructing special angles
The special angles are \latex{ 30º, 45º, 60º }, and \latex{ 90º }.
When constructing them, you can utilise the fact that angles can be bisected, copied, added, and subtracted.
Example 4
Construct a \latex{ 45º } angle.
Solution
By bisecting a \latex{ 180º } angle, you get two \latex{ 90º } angles. If you bisect the resulting angle, you get two \latex{ 45º } angles.
Sketch:
The steps of construction:
- Construct a \latex{ 180º } angle.
- Bisect the angle to get two \latex{ 90º } angles.
- Draw intersecting arcs with the same radii from the points of intersection of the arc and the arms of the \latex{ 90º } angle. Connect the resulting point of intersection and the vertex of the angle.
Construction:
\latex{ 45º }
\latex{ A }
\latex{ C }
\latex{ B }
Note:
A \latex{ 135º } angle supplements a \latex{ 45º } angle to form a \latex{ 180º } angle. So, in the previous example, a \latex{ 135º } angle was also constructed.
Example 5
Construct a \latex{ 60º } angle.
Solution
You know that every angle of an equilateral triangle is \latex{ 60º }. You already know how to construct an equilateral triangle. The construction of a \latex{ 60º } angle is similar to that of an equilateral triangle.
Sketch:
The steps of construction:
- Draw a ray with a starting point \latex{ A }. This will be one of the arms of the angle.
- Draw an arc with an arbitrary radius and centre \latex{ A }, intersecting the ray.
- Using the same radius, draw an arc with centre \latex{ B }, intersecting the previously constructed arc. Their point of intersection is \latex{ M }.
- Connect point \latex{ M } with the starting point of the ray. This will be the other arm of the angle.
Construction:
\latex{M }
\latex{ 60º }
\latex{B }
\latex{A }
Note:
By bisecting a \latex{ 60º }, you can construct a \latex{ 30º } angle. If you bisect a \latex{ 30º } angle, you can construct a \latex{ 15º } angle. A \latex{ 120º } angle can be constructed by adding two \latex{ 60º } angles.
Exercises
{{exercise_number}}. Construct a a) \latex{45º}; b) \latex{135º}; c) \latex{225º}; d) \latex{315º} angle.
{{exercise_number}}. Construct a a) \latex{60º}; b) \latex{120º}; c) \latex{240º}; d) \latex{300º} angle.
{{exercise_number}}. Construct a a) \latex{30º}; b) \latex{15º}; c) \latex{75º}; d) \latex{105º}; e) \latex{150º} angle.
{{exercise_number}}. Construct a a) \latex{22.5º}; b) \latex{67.5º}; c) \latex{112.5º}; d) \latex{157.5º} angle.
{{exercise_number}}. Construct three-fourths of the angle formed by the hands of a clock at \latex{ 3 } o'clock. How many degrees is the constructed angle?
{{exercise_number}}. Construct the face of a clock inside a circle.
{{exercise_number}}. Construct a a) 120º, b) 90º, c) 60º, d) 45º and e) 30º angle. Copy the constructed angle next to the other angles as many times as you can. Construct a circle with a radius of \latex{ 3 } \latex{ cm } around the common vertex of the angles. What type of polygons did you get in each case? Construct the line of symmetry of the resulting polygons.
{{exercise_number}}. Copy side \latex{ AB } of triangle \latex{ ABC } on a ray with starting point \latex{ D } (line segment \latex{ DE }) into your notebook. Copy angle \latex{\alpha} to point \latex{ D } and angle \latex{\beta} to point \latex{ E }. Extend the arms of the angles so that they intersect in point \latex{ F }. What can you say about the resulting \latex{ DEF } triangle? (→)
\latex{ C }
\latex{ A }
\latex{ B }
\latex{\beta}
\latex{\alpha}
{{exercise_number}}. Draw two arbitrary angles. Construct their sum. Bisect both angles and add their halves. Bisect the sum of the angles again. Compare the half of the sum of the original angles and the sum of the halves of the angles.
{{exercise_number}}. Draw a straight angle. Construct a \latex{ 135º} angle as well. Construct the sum and the difference of the angles.
{{exercise_number}}. The diagonal of a square is \latex{4.5 \; cm}. Construct the square.
{{exercise_number}}. How can you divide a \latex{90\degree} angle into three equal parts?
Quiz
An equilateral triangle was made using mirrors. A light ray enters the triangle at point \latex{ P }, forming a \latex{ 60º } angle with side \latex{ AB }. The sides of the triangle reflect the light ray. Construct the path of the light ray.
\latex{ A }
\latex{ P }
\latex{ B }
\latex{ C }
\latex{ 60° }