Euclid Book 6 Proposition 23
Statement
Given two parallelograms (ABCD, EFGH) with corresponding angles equal, the ratio of the areas of the two parallelograms is equal to the product of the ratios of their sides ().
the area of the quadrilateral ABCD
the area of the quadrilateral EFGH
ABBC
EFFG
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,F.,G.,H.},{}},{GeometricAssertion[{Parallelogram[{A.,B.,C.,D.}],Parallelogram[{E.,F.,G.,H.}]},"Similar"]},
Area[Parallelogram[{A.,B.,C.,D.}]]
Area[Parallelogram[{E.,F.,G.,H.}]]
EuclideanDistance[A.,B.]EuclideanDistance[B.,C.]
EuclideanDistance[E.,F.]EuclideanDistance[F.,G.]
Area[Parallelogram[{A.,B.,C.,D.}]] Area[Parallelogram[{E.,F.,G.,H.}]] EuclideanDistance[A.,B.]EuclideanDistance[B.,C.] EuclideanDistance[E.,F.]EuclideanDistance[F.,G.] |
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Explanations
Let , be equiangular parallelograms having the angle equal to the angle ; I say that the parallelogram has to the parallelogram the ratio compounded of the ratios of the sides.
For let them be placed so that is in a straight line with ; therefore is also in a straight line with .
Let the parallelogram be completed; let a straight line be set out, and let it be contrived that, as is to , so is to, and, as is to , so is to.
Then the ratios of to and of to are the same as the ratios of the sides, namely of to and of to . But the ratio of to is compounded of the ratio of to and of that of to ; so that has also to the ratio compounded of the ratios of the sides.
Now since, as is to , so is the parallelogram to the parallelogram ,while, as is to , so is to , therefore also, as is to , so is to .
Again, since, as is to , so is the parallelogram to ,while, as is to , so is to , therefore also, as is to , so is the parallelogram to the parallelogram .
Since then it was proved that, as is to , so is the parallelogram to the parallelogram , and, as is to, so is the parallelogram to the parallelogram , therefore, ex aequali, as is to , so is to the parallelogram . But has to the ratio compounded of the ratios of the sides; therefore also has to the ratio compounded of the ratios of the sides.
AC
CF
BCD
ECG
AC
CF
For let them be placed so that
BC
CG
DC
CE
Let the parallelogram
DG
K
BC
CG
K
L
DC
CE
L
M
Then the ratios of
K
L
L
M
BC
CG
DC
CE
K
M
K
L
L
M
K
M
Now since, as
BC
CG
AC
CH
BC
CG
K
L
K
L
AC
CH
Again, since, as
DC
CE
CH
CF
DC
CE
L
M
L
M
CH
CF
Since then it was proved that, as
K
L
AC
CH
L
M
CH
CF
K
M
AC
CF
K
M
AC
CF