PROPOSITION 25.
To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.
Let
ABC be the given rectilineal figure to which the figure to be constructed must be similar, and
D that to which it must be equal; thus it is required to construct one and the same figure similar to
ABC and equal to
D.
Let there be applied to
BC the parallelogram
BE equal to the triangle
ABC [
I. 44], and to
CE the parallelogram
CM equal to
D in the angle
FCE which is equal to the angle
CBL. [
I. 45]
Therefore
BC is in a straight line with
CF, and
LE with
EM.
Now let
GH be taken a mean proportional to
BC,
CF [
VI. 13], and on
GH let
KGH be described similar and similarly situated to
ABC. [
VI. 18]
Then, since, as
BC is to
GH, so is
GH to
CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [
VI. 19, Por.] therefore, as
BC is to
CF, so is the triangle
ABC to the triangle
KGH.
But, as
BC is to
CF, so also is the parallelogram
BE to the parallelogram
EF. [
VI. 1]
Therefore also, as the triangle
ABC is to the triangle
KGH, so is the parallelogram
BE to the parallelogram
EF; therefore, alternately, as the triangle
ABC is to the parallelogram
BE, so is the triangle
KGH to the parallelogram
EF. [
V. 16]
But the triangle
ABC is equal to the parallelogram
BE; therefore the triangle
KGH is also equal to the parallelogram
EF.
But the parallelogram
EF is equal to
D; therefore
KGH is also equal to
D.
And
KGH is also similar to
ABC.
Therefore one and the same figure
KGH has been constructed similar to the given rectilineal figure
ABC and equal to the other given figure
D. Q. E. D.
1