For, since EF has been drawn parallel to BC , one of the sides of the triangle ABC , proportionally, as BE is to EA , so is CF to FA . [[VI. 2](elem.6.2)]

Again, since FG has been drawn parallel to CD , one of the sides of the triangle ACD , proportionally, as CF is to FA , so is DG to GA . [[VI. 2](elem.6.2)]

But it was proved that, as CF is to FA , so also is BE to EA ; therefore also, as BE is to EA , so is DG to GA , and therefore, componendo , as BA is to AE , so is DA to AG , [[V. 18](elem.5.18)] and, alternately, as BA is to AD , so is EA to AG . [[V. 16](elem.5.16)]

Therefore in the parallelograms ABCD , EG , the sides about the common angle BAD are proportional.

And, since GF is parallel to DC , the angle AFG is equal to the angle DCA ; and the angle DAC is common to the two triangles ADC , AGF ; therefore the triangle ADC is equiangular with the triangle AGF .

For the same reason the triangle ACB is also equiangular with the triangle AFE , and the whole parallelogram ABCD is equiangular with the parallelogram EG .

Therefore, proportionally, as AD is to DC , so is AG to GF , as DC is to CA , so is GF to FA , as AC is to CB , so is AF to FE , and further, as CB is to BA , so is FE to EA .

And, since it was proved that, as DC is to CA , so is GF to FA , and, as AC is to CB , so is AF to FE , therefore, ex aequali , as DC is to CB , so is GF to FE . [[V. 22](elem.5.22)]

Therefore in the parallelograms ABCD , EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG . [[VI. Def. 1](elem.6.def.1)]

For the same reason the parallelogram ABCD is also similar to the parallelogram KH ; therefore each of the parallelograms EG , HK is similar to ABCD .

But figures similar to the same rectilineal figure are also similar to one another; [[VI. 21](elem.6.21)] therefore the parallelogram EG is also similar to the parallelogram HK .

Therefore etc. Q. E. D.