| To inscribe a triangle equiangular with a given triangle in a given circle. | ||
| Let ABC be the given circle, and DEF the given triangle.
It is required to inscribe a triangle equiangular with the triangle DEF in the circle ABC. | ||
| Draw GH touching the circle ABC at A. Construct the angle HAC equal to the angle DEF on the straight line AH and at the point A on it, and construct the angle GAB equal to the angle DFE on the straight line AG and at the point A on it. Join BC. | III.16,Cor
I.23 | |
| Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC equals the angle ABC in the alternate segment of the circle. | III.32 | |
| But the angle HAC equals the angle DEF, therefore the angle ABC also equals the angle DEF. | ||
| For the same reason the angle ACB also equals the angle DFE, therefore the remaining angle BAC also equals the remaining angle EDF. | I.32 | |
| Therefore a triangle equiangular with the given triangle has been inscribed in the given circle. | IV.Def.2 | |
| Q.E.F. | ||
