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Ruler and Compass Construction When we mention “geometric construction” of figures and lengths in Math, it is meant to use only ruler and compass to construct a specific figure or length. And the use of ruler and compass is governed by the following rules:
1. The ruler is used as “straightedge”. In other words, there is no mark on the ruler to indicate the length. Ruler is only used to draw straight line. 2. The compass is used for drawing circles with given radius. 3. The number of steps for using ruler and compass must be finite. It is not allowed to repeat certain process infinitely many times.
Due to these constraints, not every figure can be constructible via geometric construction approach, for example, Angle Trisection. The feasibility about Angle Trisection will be explored in Galois Theory by considering its algebraic extension field. That is out the scope of the article here. We only introduce the most basic things here.
Example ( Midpoint of line segment ) Given a line segment , find the midpoint of this line segment.
Sol: 1. For known line segment , we try to find the midpoint of it. Use point A as a center with radius larger than ½ to draw a circle. Similarly, use B as a center and the same length as radius to draw another circle. We denote the intersection of the two circles as point P and Q.
2. Link . We denote the intersection of and as point M. Then M is the midpoint of .
Proof: 1. The radii of the two circles are the same ( we draw in this way ). Thus, = , = . Furthermore, = PAQ PBQ ( SSS ) So, APQ = BPQ .
2. = , APM = BPM, = APM BPM ( SAS ) = , PMA = PMB
Since = , M is really the midpoint of the line segment.
To go a step further, PMA + PMB = 1800 and PMA = PMB PMA = PMB = 900 . is also perpendicular to ; is the perpendicular bisector of the line segment.
So, the following problems can be solved by using similar steps:
Q1: Given a line segment , find its perpendicular bisector. Q2: Given a line with a point P on it, find a line that is perpendicular to and crosses at P. Hint : Find point C and D such that P is the midpoint of that line segment.
Q3: Given a line and a point P not on , find a line that is perpendicular to and passes P. Hint: Start with drawing the following diagram
Q4: Give a line and a point P not on , find a line that is parallel to and passes P.
Example ( Bisector of an angle ) Given A, find a line that bisects A.
Sol:
1. Use A as center to draw a circle; the circle crosses the two sides on B and C respectively. 2. Use B as center to draw a circle, similarly use the same radius and C as center to draw another circle. Usually, the two circles should intersect on two points ( if not, use larger radius ). Just choose any one of the two crossing point and denote it as D.
3. Construct . Then is the bisector of the angle.
Proof:
It can be done by using SSS to show that ABD ACD . Then ADB = ADC .
Exercise: Given a circle, find the center of the circle. ( Hint: Find three points on the circle and construct perpendicular bisectors. ) |
