Draw Equilateral Triangle Inscribed in a Circle

Use but your compass and straight border when drawing a construction. No free-mitt drawing!

We will exist doing THREE constructions of an equilateral triangle. The get-go will be to construct an equilateral triangle given the length of i side, and the other two will exist to construct an equilateral triangle inscribed in a circle.

Given: the length of i side of the triangle Construct: an equilateral triangle

STEPS:
ane. Place your compass point on A and measure the distance to point B. Swing an arc of this size to a higher place (or below) the segment.
two. Without irresolute the span on the compass, place the compass bespeak on B and swing the aforementioned arc, intersecting with the outset arc.
3. Characterization the point of intersection every bit the third vertex of the equilateral triangle.

See the full circles at piece of work.

Proof of Construction: Circle A is congruent to circle B, since they were each formed using the same radius length, AB. Since AB and AC are lengths of radii of circle A, they are equal to i another. Similarly, AB and BC are radii of circle B, and are equal to i another. Therefore, AB = Air-conditioning = BC by commutation (or transitive property). Since congruent segments have equal lengths, and ΔABC is equilateral (having three congruent sides).

Given: a piece of paper Construct: an equilateral triangle inscribed in a circle.

This is a modification of the structure of a regular hexagon inscribed in a circle.

STEPS:
1. Place your compass point on the paper and depict a circle. (Proceed this compass span!)
2. Place a dot, labeled A, anywhere on the circumference of the circle to act as a starting point.
iii. Without irresolute the span on the compass, place the compass signal on A and swing a small arc crossing the circumference of the circle.
4. Without changing the span on the compass, move the compass point to the intersection of the previous arc and the circumference and brand another small arc on the circumference of the circumvolve.
5. Keep repeating this process of "stepping" around the circumvolve until you lot return to point A.
6. Starting at A, connect every other arc on the circumvolve to form the equilateral triangle.

Proof of Construction: The proof of the inscribed regular hexagon shows that the fundamental angles of a regular hexagon contain 60º. The central angles of the triangle inscribed in this circle contain 120º. Since ΔAOC is isosceles (OA and OC are radii lengths), thousand∠OCA = m∠OAC = ½ (180 - 120) = 30º. ΔAOC ΔCOB ΔBOA by SAS. By CPCTC, ∠OCB ∠OCA and m∠OCB = 30º past commutation and k∠BCA = 60º. In similar fashion, we take m∠ACB = m∠CBA = m∠BAC = 60º and equilateral ΔABC.

Given: a slice of paper Construct: an equilateral triangle inscribed in a circle.

This method uses knowledge of the special right triangle 30º - 60º - 90º.

STEPS:
1. Place your compass point on the paper and draw a circle, O. (Keep this compass span!)
2. Using a straightedge, draw a diameter of the circle, labeling the endpoints P and B.
3. Without irresolute the span on the compass, place the compass point on P and draw a full circle.
4. Label the points of intersection of the 2 circumvolve circumferences with A and C.
v. Draw segments from A to B, B to C and C to A, to form the equilateral triangle.

Proof of Construction: This construction uses the fact that an angle inscribed in a semicircle is a right angle, and that in a 30º-60º-90º triangle, the length of the short leg is half of the length of the hypotenuse. In this construction, circle O and circle P are congruent since they have the same radius length. AP is a radius length of circumvolve P and radii AP = OP. OP is also a radius length of circle O (along with OB) and diameter BP = BO + OP = 2 OP. Past commutation, BP = 2 AP, creating the weather necessary for 1000∠ABP = 30º. Consequently, m∠APB = 60º. A like argument can be used to constitute that for ΔPBC, m∠ PBC = 30º and one thousand∠BPC = 60º making ΔPBC ΔPBA past ASA (with shared side from B to P).
Now, since they are the respective sides of the two congruent triangles, making ΔABC isosceles. ∠BAC ∠BCA since the base of operations angles of an isosceles triangle are congruent.
m∠ABC = m∠ABP + k∠PBC = 30º + 30º = 60º by Angle Addition Postulate and substitution. m∠BAC + 1000∠BCA + g∠ABC = 180º because the sum of the angle measures in a triangle is 180º. Since m∠BAC + g∠BAC + 60º = 180º by commutation, we know 2m∠BAC = 120º and thou∠BAC = 60º. Consequently grand∠BCA also equals 60º by substitution, making ΔABC equilateral.

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Source: https://mathbitsnotebook.com/Geometry/Constructions/CCconstructionEqui.html

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