therefore BC and CF will be in one straight line, and LE and EM will be in one straight line.
Between BC and CF find a mean proportional GH, [VI. 13.
and on GH describe the rectilineal figure KGH, similar and similarly situated to the rectilineal figure ABC. [VI. 18.
KGH shall be the rectilineal figure required.
For, because BG is to GH as GH is to GF, [Construction.
and that if three straight lines be proportionals, as the first is to the third so is any figure on the first to a similar and similarly described figure on the second, [VI. 20, Cor. 2.
therefore as BC is to GF so is the figure ABC to the figure KGH.
But as BC is to CF so is the parallelogram BE to the parallelogram CM; [VI. 1.
therefore the figure ABG is to the figure KGH as the parallelogram BE is to the parallelogram GM. [V. 11.
And the figure ABG is equal to the parallelogram BE;
therefore the rectilineal figure KGH is equal to the parallelogram GM. [V. 14.
But the parallelogram GM is equal to the figure D; [Constr.
therefore the figure KGH is equal to the figure D, [Axiom 1.
and it is similar to the figure ABG. [Construction.
Wherefore the rectilineal figure KGH has heen de-scribed similar to the figure ABC, and equal to D. q.e.f.
PROPOSITION 26. THEOREM.
If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.
Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the common angle BAD: ABCD and AEFG shall be about the same diameter.
For, if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the
