Construction Of And Equilateral Triangle

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Construction Of An Equilateral Triangle

This article shows how to construct an equilateral triangle with compass and straightedge or ruler. An equilateral triangle is a triangle with all three sides the same length. It starts with a given line segment which is the length of each side of the preferred equilateral triangle.

It works due to the fact that the compass width is not altered between drawing each side, confirming that they are all congruent (same length). It is similar to the 60 degree angle construction, due to the fact that the interior angles of an equilateral triangle are all 60 degrees.

Printable step-by-step instructions to use in the absence of a computer

Proof

The image below is the final drawing above.

	The case	Explanation
1	PQ, PR and QR are all congruent to AB and so all have the same length	Compass width set from AB used to draw them all
2	Triangle RPQ is an equilateral triangle with the given side length AB.	All three sides congruent. See the definition of an Equilateral triangle.

- Q.E.D

This is the step-by-step, printable guide.

After you have done this	Your work ought to look like this
Begin with the line segment AB which is the length of the sides of the desired equilateral triangle.
1. Pick a point P that will be one vertex of the finished triangle.
2. Put the point of the compasses on the point A and set its drawing end to point B. The compasses are now set to the length of the sides of the finished triangle. Do not alter it from now on.
3. With the compasses' point on P, make two arcs, each more or less where the other two vertices of the triangle will be.
4. On one of the arcs, mark a point Q that will be a second vertex of the triangle. It does not matter which arc you pick, or where on the arc you draw the point.
5. Put the compasses' point on Q and draw an arc that crosses the other arc, producing point R.
6. With the use of the straightedge, draw three lines connecting the points P,Q and R.
7. Done. The triangle PQR is an equilateral triangle. Its side length is equal to the distance AB.

Constructing a 30-60-90 triangle

This section shows you how to construct (draw) a 30 60 90 degree triangle with compass and straightedge or ruler. We are given a line segment to begin, which will become the hypotenuse of a 30-60-90 right triangle. It works by linking two other constructions: A 30 degree angle, and a 60 degree angle. Due to the fact that the interior angles of a triangle always add to 180 degrees, the third angle must be 90 degrees.

Handy step-by-step instructions for you to make use of in the preparation of a lesson note or for study in the absence of a computer

Proof

	The case	Explanation
1	Angle ∠CPQ has a measure of 30°	Constructed with the use of the procedure explained in Construction of angle 30°. See that section for method and proof.
2	Angle ∠CQP has a measure of 60°	The angle is constructed with the use of the procedure described in construction of angle 60°. See that page for method and proof.
3	Angle ∠PCQ has a measure of 90°	Interior angles of a triangle add to 180°. Other two are 30° and 60° See Interior angles of a triangle.
4	PQC is a 30-60-90 triangle	(1), (2), (3)

- Q.E.D

See below the step-by-step, printable instruction to guide your learning

After you have done this	Your work ought to look like this
We begin with a line segment PQ that will be the hypotenuse of the triangle.
First, we construct a 30° angle on one end of the hypotenuse. Review the section on how to construct angle 30°.
1. Beginning with the hypotenuse line PQ, set the compasses on P, and set its width to any width that is convenient for you.
2. Draw a broad arc across PQ. Label the point where it crosses PQ as point S.
3. Without altering the compasses' width, move the compasses to the point S. Draw a broad arc that crosses the first one. Label the point where the two arcs cross as point T.
4. Again without altering the compasses' width, move the compasses to the point T, and draw an arc across the preceding arc, producing point R.
5. Draw a line from P through R and extend it beyond R to form the second side of the triangle.
After this, we construct angle 60° on the other end of the hypotenuse. See construction of angle 60°.
6. Place the compasses on point Q and set it to any suitable width.
7. Draw a broad arc across PQ on the same side as point R. Label the point where it crosses PQ as point A.
8. With the compasses on A, draw a second arc, crossing the first arc at point B
9. Draw a line from Q, through B and on to cross the line PR. Label the intersection point C.
10. You are done when you have got to this stage. The triangle PQC is a 30-60-90 triangle.

This page illustrates how to construct a right triangle that has the hypotenuse (H) and one leg (L) given. It is nearly the same construction as Perpendicular at a point on a line, apart from the fact that the compass widths used are H and L rather than the use of arbitrary widths.

Printable step-by-step instructions

The above animation is accessible as a printable step-by-step instruction manual, which can be used for producing handouts or when a computer is not accessible.

Multiple triangles possible

It is possible to draw more than one triangle has the side lengths as given. You can make use of the triangle to the left or right of the initial perpendicular, and as well draw them below the initial line. All four are correct in that they satisfy the requirements, and are congruent to each other.

Proof

This construction works by efficiently building two congruent triangles. The image below is the final drawing above with the blue line BP added

	The case or argument	Explanation to the argument
We first and foremost prove that ∆BCA is a right triangle
1	CP is congruent to CA	They were both drawn with the same compass width
2	BP is congruent to BA	They were both drawn with the same compass width
3	CB is common to both triangles BCP and BCA	Common side
4	Triangles ∆BCP and ∆BCA are congruent	Three sides congruent (SSS).
5	∠BCP, ∠BCA are congruent	CPCTC. Corresponding parts of congruent triangles are congruent
6	m∠BCA = 90°	∠BCA and ∠BCP are a linear pair and (so add to 180°) and congruent so each must be 90°
We now prove the triangle is the right size
7	CA is congruent to the given leg L	CA copied from L. See Copying a segment.
8	AB is congruent to the given hypotenuse H	Drawn with same compass width
9	∆BCA is a right triangle with the desired side lengths	From (6), (7), (8)

Below is the step-by-step, printable instruction for an easy to learn guide

	After you have done this	Your work ought to look like this
	Begin with the given segment lengths for the hypotenuse (H) and the given leg (L).
1	Draw a long horizontal line
2	Mark a point C everywhere near the middle of that line.
3	Set the compass width to the length of the given leg L
4	With the compasses on C, make an arc on each side of C, forming points P and A
5	Set the compass' width to the length of the given hypotenuse H.
6	With the compasses on P, make an arc above C
7	Without altering the compass width, move to A and make another arc over C, creating point B.
8	With the use of straight edge, draw the lines BC and BA
9	Done. ABC is a right triangle with the given leg and hypotenuse length

This page illustratess how to construct a right triangle that has the hypotenuse (H) and one angle (A) given. It works in three steps:

1. Copy the angle A. (See Copying an angle)
2. Copy the length of the hypotenuse onto the angle leg (See Copying a segment)
3. Drop a perpendicular from the end of the hypotenuse. (See Perpendicular to a line from a point)

Printable step-by-step instructions

The instruction is available in an easy to print format for handy use in creating handouts and when you study with or without a computer.

Proof

	The case or argument	Reason or explanation behind the argument
We first prove that ∆BCA is a right triangle
1	m∠BCA = 90°	BC was constructed using the procedure in Perpendicular. See that page for proof.
2	Thus ∆BCA is a right triangle	By definition of a right triangle, one angle must be 90°
Now prove BA is the hypotenuse H
3	AB = the given hypotenuse H	AB was copied from H at the same compass width
At this point prove ∠BAC is the given angle A
4	m∠BAC = given m∠A	Copied with the use of the procedure in Copying an angle. See that page for proof
9	∆BCA is a right triangle with the desired hypotenuse H and angle A	From (2), (3), (4)

- Q.E.D

The information in the table below is a step-by-step, printable instruction to guide and assist you for easy learning.

	After you have done this	Your work ought to look like this
	Start with the given hypotenuse H and angle A
1	Draw a long horizontal line
2	Mark a point A towards the left end of that line.
3	Draw an arc across the given angle A.
4	Without altering the compass width, draw a similar arc on the line at A
5	Set the compass width to the width of the given angle A
6	With the same compass width, draw an arc across the preceding one at A
7	Draw a long line from A through the point where the arcs crisscross.
8	Set the compass width to the length of the given hypotenuse H
9	Without altering the compass, draw an arc at A across the line just drawn, producing point B
10	Set the compasses on B
11	Change the compass width to be below the horizontal line
12	Draw two arcs across the horizontal line, producing points P and Q
13	With the compass at any medium width, from P and Q draw intersecting arcs under the line.
14	With the use of the straight edge, draw a line from B through where the two arcs intersect
	Done. ABC is a right triangle with the given hypotenuse H and given angle A