How to Draw a Circle Inside an Equilateral Triangle

This page shows how to construct (draw) an equilateral triangle inscribed in a circle with a compass and straightedge or ruler. This is the largest equilateral triangle that volition fit in the circle, with each vertex touching the circumvolve. This is very similar to the construction of an inscribed hexagon, except we apply every other vertex instead of all vi.

As can be seen in Definition of a Hexagon, each side of a regular hexagon is equal to the distance from the eye to any vertex. This construction simply sets the compass width to that radius, and and then steps that length off around the circle to create the 6 vertices of a hexagon.

Just instead of drawing a hexagon, we use every other vertex to brand a triangle instead. Since the hexagon construction effectively divided the circle into six equal arcs, by using every other signal, we split up information technology into 3 equal arcs instead. The three chords of these arcs course the desired equilateral triangle.

Another way of thinking about it is that both the hexagon and equilateral triangle are regular polygons, one with double the number of sides of the other.

The image below is the final drawing from the above animation, but with extra lines and the vertices labelled.

	Argument	Reason
NOTE: Steps 1 through 7 are the same as for the construction of a hexagon inscribed in a circle. In the instance of an inscribed equilateral triangle, we use every other point on the circle.
1	A,B,C,D,Due east,F all lie on the circle eye O	By construction.
2	AB = BC = CD = DE = EF	They were all fatigued with the same compass width.
From (two) nosotros see that five sides are equal in length, only the concluding side FA was not fatigued with the compasses. Information technology was the "left over" infinite as nosotros stepped around the circumvolve and stopped at F. So we accept to prove it is coinciding with the other five sides.
3	OAB is an equilateral triangle	AB was drawn with compass width set to OA, and OA = OB (both radii of the circumvolve).
iv	m∠AOB = 60°	All interior angles of an equilateral triangle are 60°.
5	m∠AOF = sixty°	Every bit in (4) m∠BOC, chiliad∠COD, k∠DOE, one thousand∠EOF are all &60deg; Since all the central angles add to 360°, m∠AOF = 360 - 5(threescore)
6	Triangle BOA, AOF are coinciding	SAS Run into Test for congruence, side-angle-side.
7	AF = AB	CPCTC - Respective Parts of Congruent Triangles are Congruent
And then now we tin can prove that BDF is an equilateral triangle
8	All six central angles (∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF, ∠FOA) are congruent	From (4) and by repetition for the other 5 angles, all six angles have a measure of 60°
ix	The angles ∠BOD, ∠DOF, ∠BOF are coinciding	From (eight) - They are each the sum of two 60° angles
10	Triangles BOD, DOF and BOF are coinciding.	The sides are all equal radii of the circumvolve, and from (9), the included angles are coinciding. Run into Examination for congruence, side-angle-side
11	BDF is an equilateral triangle.	From (10) BD, DF, FB a re congruent. CPCTC - Corresponding Parts of Congruent Triangles are Coinciding. This in turn satisfies the definition of an equilateral triangle.
12	BDF is an equilateral triangle inscribed in the given circumvolve	From (eleven) and all 3 vertices B,D,F lie on the given circle.

- Q.Due east.D

Endeavour it yourself

Click here for a printable worksheet containing 2 problems to try. When y'all go to the folio, use the browser impress command to print as many equally you wish. The printed output is non copyright.

Other constructions pages on this site

• List of printable constructions worksheets

Lines

• Introduction to constructions
• Copy a line segment
• Sum of northward line segments
• Difference of two line segments
• Perpendicular bisector of a line segment
• Perpendicular at a point on a line
• Perpendicular from a line through a point
• Perpendicular from endpoint of a ray
• Divide a segment into n equal parts
• Parallel line through a point (angle copy)
• Parallel line through a signal (rhombus)
• Parallel line through a point (translation)

Angles

• Bisecting an bending
• Copy an angle
• Construct a 30° bending
• Construct a 45° angle
• Construct a threescore° angle
• Construct a 90° bending (right angle)
• Sum of n angles
• Deviation of ii angles
• Supplementary angle
• Complementary angle
• Constructing  75°  105°  120°  135°  150° angles and more

Triangles

• Re-create a triangle
• Isosceles triangle, given base of operations and side
• Isosceles triangle, given base of operations and distance
• Isosceles triangle, given leg and apex angle
• Equilateral triangle
• xxx-60-xc triangle, given the hypotenuse
• Triangle, given three sides (sss)
• Triangle, given one side and adjacent angles (asa)
• Triangle, given two angles and non-included side (aas)
• Triangle, given ii sides and included angle (sas)
• Triangle medians
• Triangle midsegment
• Triangle altitude
• Triangle altitude (exterior example)

Right triangles

• Right Triangle, given one leg and hypotenuse (HL)
• Right Triangle, given both legs (LL)
• Correct Triangle, given hypotenuse and one bending (HA)
• Right Triangle, given 1 leg and i angle (LA)

Triangle Centers

• Triangle incenter
• Triangle circumcenter
• Triangle orthocenter
• Triangle centroid

Circles, Arcs and Ellipses

• Finding the center of a circle
• Circle given three points
• Tangent at a bespeak on the circumvolve
• Tangents through an external point
• Tangents to two circles (external)
• Tangents to two circles (internal)
• Incircle of a triangle
• Focus points of a given ellipse
• Circumcircle of a triangle

Polygons

• Square given one side
• Square inscribed in a circle
• Hexagon given one side
• Hexagon inscribed in a given circle
• Pentagon inscribed in a given circle

Non-Euclidean constructions

• Construct an ellipse with string and pins
• Detect the center of a circle with whatsoever right-angled object

(C) 2011 Copyright Math Open Reference.
All rights reserved

huberoverve.blogspot.com

Source: https://www.mathopenref.com/constinequilateral.html

Share this post