Proof: By Euclid
(related to Proposition: Prop. 11.31: Parallelepipeds on Equal Bases and Same Height are Equal in Volume)
- So, let the (straight lines) standing up, $HK$, $BE$, $AG$, $LM$, $PQ$, $DF$, $CO$, and $RS$, first of all, be at right angles to the bases $AB$ and $CD$.
- And let $RT$ have been produced in a straight line with $CR$.
- And Let (angle) $TRU$, equal to angle $ALB$, have been constructed on the straight line $RT$, at the point $R$ on it [Prop. 1.23].
- And let $RT$ be made equal to $AL$, and $RU$ to $LB$.
- And let the base $RW$, and the solid $XU$, have been completed.
- And since the two (straight lines) $TR$ and $RU$ are equal to the two (straight lines) $AL$ and $LB$ (respectively), and they contain equal angles, parallelogram $RW$ is thus equal and similar to parallelogram $HL$ [Prop. 6.14].
- And, again, since $AL$ is equal to $RT$, and $LM$ to $RS$, and they contain right angles, parallelogram $RX$ is thus equal and similar to parallelogram $AM$ [Prop. 6.14].
- So, for the same (reasons), $LE$ is also equal and similar to $SU$.
- Thus, three parallelograms of solid $AE$ are equal and similar to three parallelograms of solid $XU$.
- But, the three (faces of the former solid) are equal and similar to the three opposite (faces), and the three (faces of the latter solid) to the three opposite (faces) [Prop. 11.24].
- Thus, the whole parallelepipedal solid $AE$ is equal to the whole parallelepipedal solid $XU$ [Def. 11.10] .
- Let $DR$ and $WU$ have been drawn across, and let them have met one another at $Y$.
- And let $aTb$ have been drawn through $T$ parallel to $DY$.
- And let $PD$ have been produced to $a$.
- And let the solids $YX$ and $RI$ have been completed.
- So, solid $XY$, whose base is parallelogram $RX$, and opposite (face) $Yc$, is equal to solid $XU$, whose base (is) parallelogram $RX$, and opposite (face) $UV$.
- For they are on the same base $RX$, and (have) the same height, and the (ends of the straight lines) standing up in them, $RY$, $RU$, $Tb$, $TW$, $Se$, $Sd$, $Xc$ and $XV$, are on the same straight lines, $YW$ and $eV$ [Prop. 11.29].
- But, solid $XU$ is equal to $AE$.
- Thus, solid $XY$ is also equal to solid $AE$.
- And since parallelogram $RUWT$ is equal to parallelogram $YT$.
- For they are on the same base $RT$, and between the same parallels $RT$ and $YW$ [Prop. 1.35].
- But, $RUWT$ is equal to $CD$, since (it is) also (equal) to $AB$.
- parallelogram $YT$ is thus also equal to $CD$.
- And $DT$ is another (parallelogram).
- Thus, as base $CD$ is to $DT$, so $YT$ (is) to $DT$ [Prop. 5.7].
- And since the parallelepipedal solid $CI$ has been cut by the plane $RF$, which is parallel to the opposite planes (of $CI$), as base $CD$ is to base $DT$, so solid $CF$ (is) to solid $RI$ [Prop. 11.25].
- So, for the same (reasons), since the parallelepipedal solid $YI$ has been cut by the plane $RX$, which is parallel to the opposite planes (of $YI$), as base $YT$ is to base $TD$, so solid $YX$ (is) to solid $RI$ [Prop. 11.25].
- But, as base $CD$ (is) to $DT$, so $YT$ (is) to $DT$.
- And, thus, as solid $CF$ (is) to solid $RI$, so solid $YX$ (is) to solid $RI$.
- Thus, solids $CF$ and $YX$ each have the same ratio to $RI$ [Prop. 5.11].
- Thus, solid $CF$ is equal to solid $YX$ [Prop. 5.9].
- But, $YX$ was show (to be) equal to $AE$.
- Thus, $AE$ is also equal to $CF$.
- And so let the (straight lines) standing up, $AG$, $HK$, $BE$, $LM$, $CO$, $PQ$, $DF$, and $RS$, not be at right angles to the bases $AB$ and $CD$.
- Again, I say that solid $AE$ (is) equal to solid $CF$.
- For let $KN$, $ET$, $GU$, $MV$, $QW$, $FX$, $OY$, and $SI$ have been drawn from points $K$, $E$, $G$, $M$, $Q$, $F$, $O$, and $S$ (respectively) perpendicular to the reference plane (i.e., the plane of the bases $AB$ and $CD$), and let them have met the plane atpoints $N$, $T$, $U$, $V$, $W$, $X$, $Y$, and $I$ (respectively).
- And let $NT$, $NU$, $UV$, $TV$, $WX$, $WY$, $YI$, and $IX$ have been joined.
- So solid $KV$ is equal to solid $QI$.
- For they are on the equal bases $KM$ and $QS$, and (have) the same height, and the (straight lines) standing up in them are at right angles to their bases (see first part of proposition).
- But, solid $KV$ is equal to solid $AE$, and $QI$ to $CF$.
- For they are on the same base, and (have) the same height, and the (straight lines) standing up in them are not on the same straight lines [Prop. 11.30].
- Thus, solid $AE$ is also equal to solid $CF$.
- Thus, parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another.
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
