What are the 4 steps in constructing equilateral triangle

Draw lines from the new point to the first two points.

We have finished constructing an equilateral triangle.

In this lesson, we will learn

• how to construct a 60 degrees angle.
• how to construct an equilateral triangle.
• how to Inscribe a Regular Hexagon in a Circle.
• how to Inscribe an Equilateral Triangle in a Circle.

We now construct a 60˚ angle by constructing an equilateral triangle. Recall that the angles in an equilateral triangle are 60˚.

Go through the steps below to construct an equilateral triangle. Pick any length for a side of the equilateral triangle.

Example:

Construct

= 60˚.  

Solution:

Step 1 : Mark the point B on the line.

 

Step 2 : Stretch the compasses to any width. Put the sharp end of the compasses at point B and draw an arc on the line. Label the point where the arc intersects the line as point C.

 

Step 3: While keeping the sharp end of the compasses at point B, move the compasses away from C and draw a second arc above the line about mid-way between points B and C.

 

Step 4: Without changing the width of the compasses, place the sharp end of the compasses at point C and draw an arc to intersect the second arc.

 

Step 5 : Draw a line from point B to the point of intersection of the 2 arcs. Label the angle 60˚.

 

How to construct a 60 degree angle and an equilateral triangle?

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How to construct a 60 degree angle using only a compass and straight edge?

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How to construct an equilateral triangle?

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How to Inscribe a Regular Hexagon in a Circle?

1. Measure the distance of the circle’s radius.
   
2. Plot a point on the circle.
   
3. Starting from that point, use the compass to measure the distance of the radius and make an arc intersecting the circle.
   
4. Repeat step 3 around the circle until you return to the original point.
   
5. Connect the six points to form a hexagon.

How to Inscribe an Equilateral Triangle in a Circle?

1. Inscribe a hexagon in the circle.
   
2. Connect every other point on the hexagon to form an equilateral triangle.
   

• Show Step-by-step Solutions

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Use only your compass and straight edge when drawing a construction. No free-hand drawing!

We will be doing THREE constructions of an equilateral triangle. The first will be to construct an equilateral triangle given the length of one side, and the other two will be to construct an equilateral triangle inscribed in a circle.

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

STEPS:
1. Place your compass point on A and measure the distance to point B. Swing an arc of this size above (or below) the segment.
2. Without changing the span on the compass, place the compass point on B and swing the same arc, intersecting with the first arc.
3. Label the point of intersection as the third vertex of the equilateral triangle.

See the full circles at 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 one another. Similarly, AB and BC are radii of circle B, and are equal to one another. Therefore, AB = AC = BC by substitution (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 construction of a regular hexagon inscribed in a circle.

STEPS:
1. Place your compass point on the paper and draw a circle. (Keep this compass span!)
2. Place a dot, labeled A, anywhere on the circumference of the circle to act as a starting point.
3. Without changing the span on the compass, place the compass point 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 make another small arc on the circumference of the circle.
5. Keep repeating this process of "stepping" around the circle until you return to point A.
6. Starting at A, connect every other arc on the circle to form the equilateral triangle.

Proof of Construction: The proof of the inscribed regular hexagon shows that the central 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), m∠OCA = m∠OAC = ½ (180 - 120) = 30º. ΔAOC

ΔCOB ΔBOA by SAS. By CPCTC, ∠OCB ∠OCA and m∠OCB = 30º by substitution and m∠BCA = 60º. In similar fashion, we have m∠ACB = m∠CBA = m∠BAC = 60º and equilateral ΔABC.

Given: a piece 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 changing the span on the compass, place the compass point on P and draw a full circle.
4. Label the points of intersection of the two circle circumferences with A and C.
5. 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 circle P and radii AP = OP. OP is also a radius length of circle O (along with OB) and diameter BP = BO + OP = 2 OP. By substitution, BP = 2 AP, creating the conditions necessary for m∠ABP = 30º. Consequently, m∠APB = 60º. A similar argument can be used to establish that for ΔPBC, m∠PBC = 30º and m∠BPC = 60º making ΔPBC

ΔPBA by ASA (with shared side from B to P).
Now, since they are the corresponding sides of the two congruent triangles, making ΔABC isosceles. ∠BAC ∠BCA since the base angles of an isosceles triangle are congruent.
m∠ABC = m∠ABP + m∠PBC = 30º + 30º = 60º by Angle Addition Postulate and substitution. m∠BAC + m∠BCA + m∠ABC = 180º because the sum of the angle measures in a triangle is 180º. Since m∠BAC + m∠BAC + 60º = 180º by substitution, we know 2m∠BAC = 120º and m∠BAC = 60º. Consequently m∠BCA also equals 60º by substitution, making ΔABC equilateral.

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