| In equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles. | ||
| Let ABC and DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC equal to the angle CDE, and the angle ACB equal to the angle CED.
I say that in the triangles ABC and DEC the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles. | ||
| Let BC be placed in a straight line with CE. | ||
| Then, since the sum of the angles ABC and ACB is less than two right angles, and the angle ACB equals the angle DEC, therefore the sum of the angles ABC and DEC is less than two right angles. Therefore BA and ED, when produced, will meet. Let them be produced and meet at F. | I.17 | |
| Now, since the angle DCE equals the angle ABC, DC is parallel to FB. Again, since the angle ACB equals the angle DEC, AC is parallel to FE. | I.28 | |
| Therefore FACD is a parallelogram, therefore FA equals DC, and AC equals FD. | I.34 | |
| And, since AC is parallel to a side FE of the triangle FBE, therefore BA is to AF as BC is to CE. | ||