A kosarad üres
Special point sets
Perpendicular bisector
DEFINITION: The perpendicular bisector of a line segment in the plane is the straight line passing through the midpoint of the line segment and being perpendicular to it.
Example 1
Let us take points A and B in the plane. Let us find the points of the plane which are equidistant from point A and from point B.
Figure 55
\latex{ A }
\latex{ F }
\latex{ B }
Solution
There is only one such point on line segment AB, it is mid-point F of the line segment (Figure 55). If we draw arcs about point A and point B while adjusting the compass to longer than half of line segment AB, then we get two more suitable planar points (P and Q) (Figure 56). It can be proven that the straight line defined by points P and Q passes through point F, it is perpendicular to line segment AB, and every point of this straight line
is equidistant from points A and B. It can also be proven that there are no other suitable points in the plane apart from the points of this straight line.
is equidistant from points A and B. It can also be proven that there are no other suitable points in the plane apart from the points of this straight line.
Figure 56
\latex{ P }
\latex{ F }
\latex{ A }
\latex{ B }
\latex{ Q }
THEOREM: The set of points of the plane which are equidistant from points A and B is the perpendicular bisector of the line segment AB.
It can be proven that in the space the set of points equidistant from the given points A and B is the plane perpendicular to line segment AB and passing through its mid-point, the perpendicular bisecting plane of the line segment. (Figure 57)
Figure 57
\latex{ F }
\latex{ A }
\latex{ B }
Angle bisector
DEFINITION: The angle bisector of an angle is the ray starting from the vertex of the angle and being inside the angular domain which divides the angle into two angles with equal size.
Example 2
Let us take a convex angle in the plane. Let us find the points of the angular domain which are equidistant from the two arms of the angle.
Solution
Vertex C of the angle is a suitable point, as it lies on both arms. Let us take two straight lines which are parallel with the two arms of the angle, are equidistant from the arms and intersect each other in the angular domain. Intersection point M of these two straight lines satisfies the conditions. (Figure 58)
Figure 57
\latex{ M }
\latex{ C }
\latex{ d }
\latex{ d }
It can be proven that all the points of ray CM are equidistant from the two arms of the angle, and that no other points of the angular domain have this property. Ray CM divides the original angle into two angles of equal size.
THEOREM: The set of points in a convex angular domain which are equidistant from the arms is the angle bisector of the angle.
Example 3
Let us prove that the angle bisectors of an angle and of its adjacent angle are perpendicular to each other.
Figure 59
\latex{\alpha}
\latex{\beta}
\latex{\frac{\alpha }{2} }
\latex{\frac{\beta }{2} }
Solution
The proof using the notations of figure 59 is as follows
\latex{\alpha +\beta =180°}, therefore \latex{\frac{\alpha }{2} +\frac{\beta }{2} =90°}
Example 4
Let us construct the angle bisector of a given convex angle.
Solution
The steps of the construction can be seen in Figure 60.
⯁ ⯁ ⯁
DEFINITION: The set of points of the plane which are at a given distance r from a given point O of the plane is a circle (boundary line of the circle). (Figure 61)
DEFINITION: The set of points of the plane which are at a distance not greater than the given distance r from a given point O of the plane is a closed disc with centre O and with radius r. (Figure 62)
DEFINITION: The set of points of the plane which are at a distance less than the given distance r from a given point O of the plane is an open disc with centre O and with radius r. (Figure 63)
Figure 60
CA=CB
AM=BM
\latex{ A }
\latex{ A }
\latex{ A }
\latex{ B }
\latex{ B }
\latex{ B }
\latex{ C }
\latex{ C }
\latex{ C }
\latex{ M }
\latex{ M }
\latex{ 1. }
\latex{ 2. }
\latex{ 3. }
Figure 61
\latex{O}: centre
\latex{r}: radius
\latex{OP = r}
\latex{r}: radius
\latex{OP = r}
(BOUNDARY
OF THE)
CIRCLE
OF THE)
CIRCLE
\latex{ O }
\latex{ r }
\latex{ P }
Figure 62
OP\latex{\leq }r
CLOSED
DISC
DISC
\latex{ O }
\latex{ r }
\latex{ P }
Figure 62
\latex{OP} \latex{\lt }r
OPEN
DISC
DISC
\latex{ O }
\latex{ P }
\latex{ r }
Figure 64
CHORD
DIAMETER
SECANT LINE
SEGMENT
SECTOR
ARC
(circles with common centre) concentric circles
RING OR ANNULUS
Figure 65
point of tangency
TANGENT LINE
\latex{ O }
\latex{ r }
The designations relating to a circle and its parts are shown in Figure 64. A circle and a straight line can have \latex{ 0 }, \latex{ 1 } or \latex{ 2 } points in common.
DEFINITION: A tangent of a circle is a straight line in the plane of the circle which has exactly one point in common with the circle. (Figure 65)
It can be proven that exactly one tangent line can be drawn at each point of a circle, and that the radius drawn to the point of tangency is perpendicular to the tangent line.
Two coplanar circles can have \latex{ 0 }, \latex{ 1 } or \latex{ 2 } points in common. If there is \latex{ 1 } point in common, then the two circles touch each other.
It can be proven that in the case of touching circles the centres of the circles and the point of touch are collinear. (Figure 66)
Figure 66
\latex{ O_2 }
\latex{ O_1 }
\latex{ O_1 }
\latex{ O_2 }
\latex{ E }
\latex{ E }
DEFINITION: The set of points of the space which are at a given distance from a given point is a sphere (boundary surface of the sphere).
The given point is the centre of the sphere, the given distance is the radius of sphere.
Example 5
Let us construct a tangent line at a given point of a given circle.
Solution
Given: O; r; P. The straight line perpendicular to straight line OP at point P is the tangent line we are looking for. The steps of the construction can be seen in Figure 67.
Figure 67
\latex{KP=PL}
\latex{KM=ML}
\latex{ O }
\latex{ P }
\latex{ O }
\latex{ K }
\latex{ P }
\latex{ L }
\latex{ O }
\latex{ K }
\latex{ P }
\latex{ L }
\latex{ O }
\latex{ K }
\latex{ P }
\latex{ L }
\latex{ M }
\latex{ M }
\latex{ e }
\latex{ 4. }
\latex{ 3. }
\latex{ 2. }
\latex{ 1. }
Example 6
Let us prove that the perpendicular bisector of any chord of the circle passes through the centre of the circle.
Solution
Points A and B are equidistant from centre O of the circle, therefore the centre lies on the perpendicular bisector of chord AB. (Figure 68)
Figure 68
\latex{ A }
\latex{ B }
\latex{ F }
\latex{ O }
Exercises
Do the construction of exercises 2., 3., 4., 5., 8., 10 and 11 with the help of a computer based geometric construction software too. (E.g. www.euklides.hu or other similar application.)
{{exercise_number}}. What is the included angle of the angle bisectors of two interior angles on one leg of a trapezium?
{{exercise_number}}. Take a circle with a radius of \latex{ 3\, cm }, and draw one of its chords. Construct the points of the circle which are equidistant from the two end-points of the chord.
{{exercise_number}}. Take a circle and points A and B not on the circle. Construct the points of the circle which are equidistant from point A and from point B. Is there always a solution?
{{exercise_number}}. Take points A and B in the plane. Where are those points P of the plane for which the following is true:
- \latex{AP\geq BP};
- \latex{AP\lt BP?}
{{exercise_number}}. Take a \latex{ 60º } angle. Construct a circle with \latex{ 2\, cm } radius inside the angular domain so that the circle touches the arms of the angle.
{{exercise_number}}. Take a 45º angle. Construct a circle with \latex{ 2\, cm } radius inside the angular domain so that the circle touches one of the arms of the angle and also touches the angle bisector. How many solutions does the exercise have?
{{exercise_number}}. Take a rectangle which is not a square. What planar figure is defined by the angle bisectors of the interior angles of the rectangle? Justify your answer.
{{exercise_number}}. Let the distance between two points A and B be \latex{ 6\, cm }. Construct a circle with
- \latex{ 2\, cm };
- \latex{ 3\, cm };
- \latex{ 4\, cm }
radius which touches both the circle with \latex{ 2\, cm } radius about point A and the circle with \latex{ 2\, cm } radius about B. How many solutions does the exercise have in each case?
{{exercise_number}}. What is the connection between the coordinates of those points of the Cartesian coordinate system which are equidistant from the coordinate axes?
{{exercise_number}}. Construct the interior angle bisectors of a given triangle. What do you experience?
{{exercise_number}}. Construct the perpendicular bisectors of the sides of a given triangle. What do you experience?
Quiz
Two points are given in the plane. Using only a pair of compasses construct two points the distance between which is twice the distance between the two given points.
