:500px Let $AB$ and $CD$ be equal parallelepipeds. It is to be demonstrated that the bases of $AB$ and $CD$ are inversely proportional to their heights. That is: :$EH : NQ = CM : AG$ First, let $AG, EF, LB, HK, CM, NO, PD, QR$ be perpendicular to their bases. It is to be shown that: :$EH : NQ = CM : AG$ Suppose $EH = NQ$. From {{EuclidPropLink|book = XI|prop = 32|title = Parallelepipeds of Same Height have Volume Proportional to Bases}}: :$CM = AG$ Thus it follows trivially that: :$EH : NQ = CM : AG$ Next, suppose $EH \ne NQ$. Suppose WLOG that $EH > NQ$. We have that $AB = CD$. Therefore $CM > AG$. Let $CT = AG$. Let the parallelepiped $CV$ be completed on the base $NQ$ and with height $CT$. We have that $AB = CD$. So from {{EuclidPropLink|book = V|prop = 7|title = Ratios of Equal Magnitudes}}: :$AB : CV = CD : CV$ But from {{EuclidPropLink|book = XI|prop = 32|title = Parallelepipeds of Same Height have Volume Proportional to Bases}}: :$AB : CV = EH : NQ$ From {{EuclidPropLink|book = XI|prop = 25|title = Parallelepiped cut by Plane Parallel to Opposite Planes}}: :$CD : CV = MQ : TQ$ and from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$CD : CV = CM : CT$ Therefore also: :$EH : NQ = MC : CT$ But: :$CT = AG$ Therefore also: :$EH : NQ = MC : AG$ Therefore in the parallelepipeds $AB$ and $CD$, the bases are inversely proportional to their heights. {{qed|lemma}} Now let the bases of the parallelepipeds $AB$ and $CD$ be inversely proportional to their heights. That is: :$EH : NQ = MC : AG$ It is to be shown that $AB = CD$. Let the sides $AG, EF, LB, HK, CM, NO, PD, QR$ be perpendicular to their bases. Let $EH = NQ$. Then it follows from $EH : NQ = MC : AG$ that $MC = AG$. From {{EuclidPropLink|book = XI|prop = 31|title = Parallelepipeds on Equal Bases and Same Height are Equal in Volume}}: :$AB = CD$ Next suppose that $EH \ne NQ$. Suppose WLOG that $EH > NQ$. Therefore the height of $CD$ is greater than the height of $AB$. That is: :$CM > AG$ Let $CT = AG$. Let the parallelepiped $CV$ be completed on the base $NQ$ and with height $CT$. We have that: :$EH : NQ = MC : AG$ while: :$AG = CT$ Therefore: :$EH : NQ = CM : CT$ We have that $AB$ and $CV$ are of equal height. So from {{EuclidPropLink|book = XI|prop = 32|title = Parallelepipeds of Same Height have Volume Proportional to Bases}}: :$EH : NQ = AB : CV$ From {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$CM : CT = MQ : QT$ and from {{EuclidPropLink|book = XI|prop = 25|title = Parallelepiped cut by Plane Parallel to Opposite Planes}}: :$CM : CT = CD : CV$ Therefore also: :$AB : CV = CD : CV$ Therefore from {{EuclidPropLink|book = V|prop = 9|title = Magnitudes with Same Ratios are Equal}}: :$AB = CD$ {{qed|lemma}} :500px Now let $AG, EF, LB, HK, CM, NO, PD, QR$ ''not'' be perpendicular to their bases. Let perpendiculars be drawn from $F, G, B, K, O, M, D, R$ to the planes through $EH$ and $NQ$, meeting them at $S, T, U, V, W, X, Y, A'$. Let the parallelepipeds $FV$ and $OA'$ be completed. We have that: :$AB = CD$ We have that $AB$ and $BT$ are on the same base $FK$ and have the same height. From: :{{EuclidPropLink|book = XI|prop = 29|title =  Parallelepipeds on Same Base and Same Height whose Extremities are on Same Lines are Equal in Volume}} and: :{{EuclidPropLink|book = XI|prop = 30|title =  Parallelepipeds on Same Base and Same Height whose Extremities are not on Same Lines are Equal in Volume}}: we have that: :$AB = BT$ For the same reason: :$CD = DX$ Therefore: :$BT = DX$ Therefore from the first part of this result: :$FK : OR$ is the same as the ratio of the heights of $DX$ and $BT$. But: :$FK = EH$ and: :$OR = NQ$ Therefore: :$EH : NQ$ is the same as the ratio of the heights of $DX$ and $BT$. But $DX$ and $BT$ have the same heights respectively as $DC$ and $BA$. Therefore: :$EH : NQ$ is the same as the ratio of the heights of $DC$ and $BA$. Therefore in the parallelepipeds $AB$ and $CD$, the bases are inversely proportional to their heights. {{qed|lemma}} Now let the bases of the parallelepipeds $AB$ and $CD$ be inversely proportional to their heights. Let perpendiculars be drawn from $F, G, B, K, O, M, D, R$ to the planes through $EH$ and $NQ$, meeting them at $S, T, U, V, W, X, Y, A'$. Let the parallelepipeds $FV$ and $OA'$ be completed. Then as $DY$ and $BU$ are the heights of $AB$ and $CD$: :$EH : NQ = DY : BU$ Thus: :$EH : NQ$ equals the ratio of the heights of $DC$ and $BA$. We have: :$EH = FK$ and: :$NQ = OR$ Therefore: :$FK : OR$ equals the ratio of the heights of $DC$ and $BA$. But $AB$ and $CD$ have the same heights as $BT$ and $DX$ respectively. Therefore: :$FK : OR$ equals the ratio of the heights of $DX$ and $BT$. Therefore the bases of the parallelepipeds $BT$ and $DX$ are inversely proportional to their heights. Therefore by the first part of this result: :$BT = DX$ But from: :{{EuclidPropLink|book = XI|prop = 29|title =  Parallelepipeds on Same Base and Same Height whose Extremities are on Same Lines are Equal in Volume}} and: :{{EuclidPropLink|book = XI|prop = 30|title =  Parallelepipeds on Same Base and Same Height whose Extremities are not on Same Lines are Equal in Volume}}: we have that: :$AB = BT$ and: :$DX = DC$ Therefore: :$AB = CD$ {{qed}} {{Euclid Note|34|XI}}