By combining the constructions of perpendicular bisectors, perpendicular lines, angle bisectors, and tangents learned so far, we can construct various figures. Let us look at some typical applications.
Circumcenter of a Triangle (Center of the Circumcircle)
The perpendicular bisectors of the three sides of a triangle always intersect at one point. This point is called the circumcenter. A circle passing through the three vertices of the triangle can be drawn with the circumcenter as its center (the circumcircle).
Position of the circumcenter: The circumcenter is equidistant from the three vertices of the triangle.
Incenter of a Triangle (Center of the Incircle)
The angle bisectors of the three interior angles of a triangle always intersect at one point. This point is called the incenter. A circle tangent to all three sides of the triangle can be drawn with the incenter as its center (the incircle).
Position of the incenter: The incenter is equidistant from the three sides of the triangle.
State the procedure for constructing the circumcircle of triangle ABC.
Solution:
① Construct the perpendicular bisector of side AB.
② Construct the perpendicular bisector of side BC.
③ The intersection O of ① and ② is the circumcenter (the center of the circumcircle).
④ Draw a circle centered at O with the distance from O to A (or B or C) as its radius → the circumcircle.
Constructing Angles of 60°, 30°, and 45°
Special angles can be constructed with a compass without using a protractor.
・Construct an equilateral triangle → each interior angle is 60°
・Bisect 60° → 30°
・Bisect a right angle → 45°
State the procedure for constructing a 60° angle with a compass.
Solution:
① Draw ray OA.
② Draw an arc with O as the center and an appropriate radius, and let its intersection with OA be P.
③ With the same radius and P as the center, draw an arc and let its intersection with the arc in ② be Q.
④ Draw ray OQ → ∠AOQ = 60° (because triangle OPQ is equilateral).
Application: A Point on a Given Line ℓ Equidistant from Two Points A and B
“Equidistant from A and B” → The point lies on the perpendicular bisector of segment AB.
“On line ℓ” → The point lies on ℓ.
The point satisfying both conditions is the intersection of the perpendicular bisector of AB and ℓ.
- Circumcenter = intersection of the perpendicular bisectors of the three sides (equidistant from the three vertices · center of the circumcircle)
- Incenter = intersection of the three angle bisectors (equidistant from the three sides · center of the incircle)
- 60°: equilateral triangle (two circles with radius equal to one side) ／ 30°: bisect 60° ／ 45°: bisect a right angle
- When you see “equidistant,” consider a perpendicular bisector or an angle bisector.
Practice Problems
- How do you construct the center of the incircle of triangle ABC?
- Describe the procedure for constructing a 30° angle using only a compass and ruler.
- Given two points A and B and line ℓ, state the procedure for constructing point P on ℓ that is equidistant from A and B.
Answers and Explanations
- Answer: Draw the three angle bisectors; their intersection is the incenter.
Explanation: The three angle bisectors intersect at one point, the incenter. - Answer: Construct a 60° angle and bisect it with an angle bisector.
Explanation: Half of 60° (from an equilateral triangle) is 30°. - Answer: Construct the perpendicular bisector of segment AB; its intersection with line ℓ is P.
Explanation: Every point on the perpendicular bisector of AB is equidistant from A and B.