|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.