Start by drawing a circle and its diameter, using a compass and a straight edge. Then, without changing the span of the compass construct a second circle, with the center in one of the endpoints of the diameter.
From here we will draw △ ABD. Notice that BD is a radius of ⊙ B. Since we did not change the span of the compass, ⊙ C and ⊙ B have the same radius.
BD=BC=AC
Therefore, we can conclude that AB is twice BD. Because one of the sides of △ ABD is a diameter of the circle, we know that m∠ BDA = 90 ^(∘) by the Inscribed Right Triangle Theorem. With this information, we can state that △ ABD is a 30 ^(∘) - 60 ^(∘) - 90 ^(∘) triangle.
Let's now connect points A, B, and E to draw a second triangle. Notice that we can use the same method as before, to conclude that obtained △ ABE is also 30 ^(∘) - 60 ^(∘) - 90 ^(∘) triangle.
Using the Hypotenuse Leg Theorem, we can also conclude that △ ABD and △ ABE are congruent triangles.
Now we will draw a segment connecting the points D and E. This way we obtain △ AED. Knowing the fact that the corresponding parts of congruent triangles are congruent, we can deduce that AD=AE. Thus, △ AED is an isosceles triangle with the vertex angle of m∠ DAE = 60 ^(∘).
Let's now focus on △ ADE. Since this is an isosceles triangle with m∠ DAE = 60 ^(∘), we can find the measure of its base angles using the Triangle Angle Sum Theorem.
m∠ ADE + m∠ DEA + 60 ^(∘) = 180 ^(∘)
⇓
m∠ ADE =m∠ DEA = 60 ^(∘)
Finally, we obtained a triangle with all angles measuring 60 ^(∘). Therefore, the constructed triangle AED is equilateral.
