PROPOSITION 28.
To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.
Let
AB be the given straight line,
C the given rectilineal figure to which the figure to be applied to
AB is required to be equal, not being greater than the parallelogram described on the half of
AB and similar to the defect, and
D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line
AB a parallelogram equal to the given rectilineal figure
C and deficient by a parallelogrammic figure which is similar to
D.
Let
AB be bisected at the point
E, and on
EB let
EBFG be described similar and similarly situated to
D; [
VI. 18] let the parallelogram
AG be completed.
If then
AG is equal to
C, that which was enjoined will have been done;
for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.
But, if not, let
HE be greater than
C.
Now
HE is equal to
GB;
therefore GB is also greater than C.
Let
KLMN be constructed at once equal to the excess by which
GB is greater than
C and similar and similarly situated to
D. [
VI. 25]
But
D is similar to
GB;
therefore KM is also similar to GB. [VI. 21]
Let, then,
KL correspond to
GE, and
LM to
GF.
Now, since
GB is equal to
C,
KM,
therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM.
Let
GO be made equal to
KL, and
GP equal to
LM; and let the parallelogram
OGPQ be completed;
therefore it is equal and similar to KM.
Therefore
GQ is also similar to
GB; [
VI. 21] therefore
GQ is about the same diameter with
GB. [
VI. 26]
Let
GQB be their diameter, and let the figure be described.
Then, since
BG is equal to
C,
KM, and in them
GQ is equal to
KM, therefore the remainder, the gnomon
UWV, is equal to the remainder
C.
And, since
PR is equal to
OS,
let QB be added to each; therefore the whole
PB is equal to the whole
OB.
But
OB is equal to
TE, since the side
AE is also equal to the side
EB; [
I. 36]
therefore TE is also equal to PB.
Let
OS be added to each;
therefore the whole TS is equal to the whole, the gnomon VWU.
But the gnomon
VWU was proved equal to
C;
therefore TS is also equal to C.
Therefore to the given straight line
AB there has been applied the parallelogram
ST equal to the given rectilineal figure
C and deficient by a parallelogrammic figure
QB which is similar to
D. Q. E. F.