Then the ratios of
K to
L and of
L to
M are the same
as the ratios of the sides, namely of
BC to
CG and of
DC to
CE.
But the ratio of
K to
M is compounded of the ratio of
K to
L and of that of
L to
M; so that
K has also to
M the ratio compounded of the ratios
of the sides.
Now since, as
BC is to
CG, so is the parallelogram
AC to the parallelogram
CH, [
VI. 1] while, as
BC is to
CG, so is
K to
L, therefore also, as
K is to
L, so is
AC to
CH. [
V. 11]
Again, since, as
DC is to
CE, so is the parallelogram
CH to
CF, [
VI. 1] while, as
DC is to
CE, so is
L to
M, therefore also, as
L is to
M, so is the parallelogram
CH to the parallelogram
CF. [
V. 11]
Since then it was proved that, as
K is to
L, so is the parallelogram
AC to the parallelogram
CH, and, as
L is to
M, so is the parallelogram
CH to the parallelogram
CF, therefore,
ex aequali, as
K is to
M, so is
AC to the parallelogram
CF.
But
K has to
M the ratio compounded of the ratios of the sides;
therefore AC also has to CF the ratio compounded of the ratios of the sides.
Therefore etc. Q. E. D.
1
2
PROPOSITION 24.
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. [VI. 2]
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, [V. 18] and, alternately,
as BA is to AD, so is EA to AG. [V. 16]
Therefore in the parallelograms
ABCD,
EG, the sides about the common angle
BAD are proportional.
And, since
GF is parallel to
DC,
