About a given circle to circumscribe a triangle equiangular with a given triangle.
Proof
Let ABC be the given circle with centre K and DEF the given
triangle. Produce EF both ways to G, H. At the centre K
construct ∠AKB=∠DEG and ∠AKC=∠DFH
(I.23). At A, B, C draw the tangents to the circle (III.16,
III.17); they bound a triangle. Because each tangent is perpendicular
to the radius at the point of contact (III.18), the angles of the
constructed triangle are the supplements of the central angles ∠AKB, ∠AKC, ∠BKC, hence equal to the angles of
△DEF by I.13 and the construction.