◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-11-elementary-stereometry / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 11.33: Volumes of Similar Parallelepipeds are in Triplicate Ratio to Length of Corresponding Sides)

• Let $AB$ and $CD$ be similar parallelepiped solids, and let $AE$ correspond to $CF$.
• I say that solid $AB$ has to solid $CD$ the cubed ratio that $AE$ (has) to $CF$.

• For let $EK$, $EL$, and $EM$ have been produced in a straight line with $AE$, $GE$, and $HE$ (respectively).
• And let $EK$ be made equal to $CF$, and $EL$ equal to $FN$, and, further, $EM$ equal to $FR$.
• And let the parallelogram $KL$ have been completed, and the solid $KP$.
• And since the two (straight lines) $KE$ and $EL$ are equal to the two (straight lines) $CF$ and $FN$, but angle $KEL$ is also equal to angle $CFN$, inasmuch as $AEG$ is also equal to $CFN$, on account of the similarity of the solids $AB$ and $CD$, parallelogram $KL$ is thus equal [and similar to parallelogram $CN$.
• So, for the same (reasons), parallelogram $KM$ is also equal and similar to [parallelogram] $CR$, and, further, $EP$ to $DF$.
• Thus, three parallelograms of solid $KP$ are equal and similar to three parallelograms of solid $CD$.
• But the three (former parallelograms) are equal and similar to the three opposite (parallelograms), and the three (latter parallelograms) are equal and similar to the three opposite (parallelograms) [Prop. 11.24].
• Thus, the whole of solid $KP$ is equal and similar to the whole of solid $CD$ [Def. 11.10] .
• Let parallelogram $GK$ have been completed.
• And let the solids $EO$ and $LQ$, with bases the parallelograms $GK$ and $KL$ (respectively), and with the same height as $AB$, have been completed.
• And since, on account of the similarity of solids $AB$ and $CD$, as $AE$ is to $CF$, so $EG$ (is) to $FN$, and $EH$ to $FR$ [Def. 6.1] , [Def. 11.9] , and $CF$ (is) equal to $EK$, and $FN$ to $EL$, and $FR$ to $EM$, thus as $AE$ is to $EK$, so $GE$ (is) to $EL$, and $HE$ to $EM$.
• But, as $AE$ (is) to $EK$, so [parallelogram] $AG$ (is) to parallelogram $GK$, and as $GE$ (is) to $EL$, so $GK$ (is) to $KL$, and as $HE$ (is) to $EM$, so $QE$ (is) to $KM$ [Prop. 6.1].
• And thus as parallelogram $AG$ (is) to $GK$, so $GK$ (is) to $KL$, and $QE$ (is) to $KM$.
• But, as $AG$ (is) to $GK$, so solid $AB$ (is) to solid $EO$, and as $GK$ (is) to $KL$, so solid $OE$ (is) to solid $QL$, and as $QE$ (is) to $KM$, so solid $QL$ (is) to solid $KP$ [Prop. 11.32].
• And, thus, as solid $AB$ is to $EO$, so $EO$ (is) to $QL$, and $QL$ to $KP$.
• And if four magnitudes are in continued proportion then the first has to the fourth the cubed ratio that (it has) to the second [Def. 5.10] .
• Thus, solid $AB$ has to $KP$ the cubed ratio which $AB$ (has) to $EO$.
• But, as $AB$ (is) to $EO$, so parallelogram $AG$ (is) to $GK$, and the straight line $AE$ to $EK$ [Prop. 6.1].
• Hence, solid $AB$ also has to $KP$ the cubed ratio that $AE$ (has) to $EK$.
• And solid $KP$ (is) equal to solid $CD$, and straight line $EK$ to $CF$.
• Thus, solid $AB$ also has to solid $CD$ the cubed ratio which its corresponding side $AE$ (has) to the corresponding side $CF$.

• Thus, similar parallelepiped solids are to one another as the cubed ratio of their corresponding sides.
• (Which is) the very thing it was required to show.
  ∎

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:

@Fitzpatrick

References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"