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.
PROPOSITION 29.
To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.
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, and
D that to which the excess is required to be similar; thus it is required to apply to the straight line
AB a parallelogram equal to the rectilineal figure
C and exceeding by a parallelogrammic figure similar to
D.
Let
AB be bisected at
E; let there be described on
EB the parallelogram
BF similar and similarly situated to
D; and let
GH be constructed at once equal to the sum of
BF,
C and similar and similarly situated to
D. [
VI. 25]
Let
KH correspond to
FL and
KG to
FE.
Now, since
GH is greater than
FB, therefore
KH is also greater than
FL, and
KG than
FE.