In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another.
Let ABCD be a parallelogram, and AC its diameter,
and let EG, HK be parallelograms about AC;
I say that each of the parallelograms EG, HK is similar both to the whole ABCD
and to the other.
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]
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.
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,
and, alternately, as BA is to AD, so is EA to AG.
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.
Therefore in the parallelograms ABCD, EG the sides about the equal angles are
proportional; therefore the parallelogram ABCD is similar to the parallelogram
EG.
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; therefore the parallelogram EG is also similar to the parallelogram
HK.
Therefore, etc.
Q.E.D.