Let DF be joined, and on the straight line AB , and at the points A , B on it, let the angle GAB be constructed equal to the angle at C , and the angle ABG equal to the angle CDF . [[I. 23](elem.1.23)]

Therefore the remaining angle CFD is equal to the angle AGB ; [[I. 32](elem.1.32)] therefore the triangle FCD is equiangular with the triangle GAB .

Therefore, proportionally, as FD is to GB , so is FC to GA , and CD to AB .

Again, on the straight line BG , and at the points B , G on it, let the angle BGH be constructed equal to the angle DFE , and the angle GBH equal to the angle FDE . [[I. 23](elem.1.23)]

Therefore the remaining angle at E is equal to the remaining angle at H ; [[I. 32](elem.1.32)] therefore the triangle FDE is equiangular with the triangle GBH ; therefore, proportionally, as FD is to GB , so is FE to GH , and ED to HB . [[VI. 4](elem.6.4)]

But it was also proved that, as FD is to GB , so is FC to GA , and CD to AB ; therefore also, as FC is to AG , so is CD to AB , and FE to GH , and further ED to HB .

And, since the angle CFD is equal to the angle AGB , and the angle DFE to the angle BGH , therefore the whole angle CFE is equal to the whole angle AGH .

For the same reason the angle CDE is also equal to the angle ABH .

And the angle at C is also equal to the angle at A , and the angle at E to the angle at H .

Therefore AH is equiangular with CE ; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE . [[VI. Def. 1](elem.6.def.1)]

Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE . Q. E. F.