Equilateral triangle inscribed in a circle
Geometry construction using a compass and straightedge
This page shows how to construct (draw) an
inscribed in a circle with a compass and straightedge or ruler. This is the largest equilateral triangle that will fit in the circle, with each
touching the circle. This is very similar to the construction of an
inscribed hexagon, except we use every other vertex instead of all six.
Printable step-by-step instructions
The above animation is available as a
printable step-by-step instruction sheet, which can be used for making handouts
or when a computer is not available.
Explanation of method
As can be seen in Definition of a Hexagon,
each side of a regular hexagon is equal to the distance from the center to any vertex.
This construction simply sets the compass width to that radius, and then steps that length off around the circle
to create the six vertices of a hexagon.
But instead of drawing a hexagon, we use every other vertex to make a triangle instead. Since the hexagon construction effectively divided the
circle into six equal arcs, by using every other point, we divide it into three equal arcs instead. The three chords of these arcs form 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.
|NOTE: Steps 1 through 7 are the same as for the construction of a hexagon inscribed in a circle.
In the case of an inscribed equilateral triangle, we use every other point on the circle.
||A,B,C,D,E,F all lie on the circle center O
||AB = BC = CD = DE = EF
||They were all drawn with the same compass width.
|From (2) we see that five sides are equal in length, but the last side FA was not drawn with the compasses.
It was the "left over" space as we stepped around the circle and stopped at F.
So we have to prove it is congruent with the other five sides.
||OAB is an equilateral triangle
||AB was drawn with compass width set to OA,
and OA = OB (both radii of the circle).
||m∠AOB = 60°
||All interior angles of an equilateral triangle are 60°.
||m∠AOF = 60°
||As in (4) m∠BOC, m∠COD, m∠DOE, m∠EOF are all &60deg;
Since all the central angles add to 360°,
m∠AOF = 360 - 5(60)
||Triangle BOA, AOF are congruent
||SAS See Test for congruence, side-angle-side.
||AF = AB
||CPCTC - Corresponding Parts of Congruent Triangles are Congruent
|So now we can prove that BDF is an equilateral triangle
||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°
||The angles ∠BOD, ∠DOF, ∠BOF are congruent
||From (8) - They are each the sum of two 60° angles
||Triangles BOD, DOF and BOF are congruent.
||The sides are all equal radii of the circle, and from (9), the included angles are congruent.
See Test for congruence, side-angle-side
||BDF is an equilateral triangle.
||From (10) BD, DF, FB a re congruent. CPCTC - Corresponding Parts of Congruent Triangles are Congruent.
This in turn satisfies the definition of an
||BDF is an equilateral triangle inscribed in the given circle
From (11) and all three vertices B,D,F lie on the given circle.
Try it yourself
Click here for a printable worksheet containing two problems to try.
When you get to the page, use the browser print command to print as many as you wish. The printed output is not copyright.
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Other constructions pages on this site
Circles, Arcs and Ellipses
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