|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
|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|
Next proposition: IV.3