:600px Let $AE$ and $CF$ be parallelepipeds of the same height on equal bases $AB$ and $CD$. It is to be demonstrated that the parallelepiped $AE$ is equal to the parallelepiped $CF$. Let the sides $HK, BE, AG, LM, PQ, DF, CO, RS$ be perpendicular to the bases $AB$ and $CD$. Let the straight line $RT$ be produced in a straight line with $CR$. From {{EuclidPropLink|book = I|prop = 23|title = Construction of Equal Angle}}: :let $\angle TRU$ be constructed on $RT$ at $R$ equal to $\angle ALB$. Let $RT = AL$. Let $RU = LB$. Let the base $RW$ and the parallelepiped $XU$ be completed. We have that the two sides $TR$ and $RU$ are equal to the two sides $AL$ and $LB$. $TR, RU$ and $AL, LB$ contain equal angles. Therefore the parallelogram $RW$ is equal and similar to the parallelogram $HL$. We have that $AL = RT$ and $LM = RS$. $AL, LM$ and $RT, RS$ contain right angles. Therefore the parallelogram $RX$ is equal and similar to the parallelogram $AM$. For the same reason, the parallelogram $LE$ is equal and similar to the parallelogram $SU$. Therefore three parallelograms of the parallelepiped $AE$ are equal and similar to three parallelograms of the parallelepiped $XU$. But from {{EuclidPropLink|book = XI|prop = 24|title = Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms}}: :the former three are equal and similar to the three opposite. Therefore from {{EuclidDefLink|XI|10|Similar Equal Solid Figures}}: :the whole parallelepiped $AE$ is equal to the whole parallelepiped $XU$. Let $DR$ and $WU$ be drawn through and meet one another at $Y$. Let $A'TB'$ be drawn through $T$ parallel to $DY$. Let $PD$ be produced to $A'$. Let the parallelepipeds $YX$ and $RI$ be completed. The parallelepiped $YX$ has base $RX$ whose opposite is $YC'$. Then $YX$ equals the parallelepiped $XU$ which has a base $RX$ whose opposite is $UV$. It follows from {{EuclidPropLink|book = XI|prop = 29|title = Parallelepipeds on Same Base and Same Height whose Extremities are on Same Lines are Equal in Volume}}: :the parallelepiped $XU$ equals the parallelepiped $XY$. But $XU = AE$. Therefore the parallelepiped $XY$ equals the parallelepiped $AE$. We have that the parallelogram $RUWT$ is on the same base as the parallelogram $YT$. Also, $RUWT$ is between the same parallels as $YT$. Therefore from {{EuclidPropLink|book = I|prop = 35|title = Parallelograms with Same Base and Same Height have Equal Area}}: :the parallelogram $RUWT$ equals the parallelogram $YT$. But $RUWT = AB$. Therefore $RUWT = CD$ Therefore $YT = CD$. But $DT$ is another parallelogram. Therefore from {{EuclidPropLink|book = V|prop = 7|title = Ratios of Equal Magnitudes}}: :as the base $CD$ is to $DT$, so is $YT$ to $DT$. We have that the parallelepiped $CI$ has been cut by the plane $RF$ which is parallel to opposite planes. So from {{EuclidPropLink|book = XI|prop = 25|title = Parallelepiped cut by Plane Parallel to Opposite Planes}}: :as the base $CD$ is to $DT$, so is the parallelepiped $CF$ to the parallelepiped $RI$. For the same reason, as the parallelepiped $YI$ has been cut by the plane $RX$ which is parallel to opposite planes, from {{EuclidPropLink|book = XI|prop = 25|title = Parallelepiped cut by Plane Parallel to Opposite Planes}}: :as the base $YT$ is to $TD$, so is the parallelepiped $YX$ to the parallelepiped $RI$. But as the base $CD$ is to $DT$, so is $YT$ is to $TD$. Therefore, also, as the parallelepiped $CF$ is to the parallelepiped $RI$, so is the parallelepiped $YX$ to $RI$. Therefore each of the parallelepipeds $CF$ and $YX$ has to $RI$ the same ratio. Therefore the parallelepiped $CF$ equals the parallelepiped $YX$. But parallelepiped $YX$ was proved equal to $AE$. Therefore $AE = CF$. {{qed|lemma}} :600px Next, suppose the sides $AG, HK, BE, LM, CN, PQ, DF, RS$ ''not'' be perpendicular to the bases $AB$ and $CD$. It is to be demonstrated that the parallelepiped $AE$ is equal to the parallelepiped $CF$. From the points $K, E, G, M, Q, F, N, S$ let $KO, ET, GU, MV, QW, FX, NY, SI$ be drawn perpendicular to the plane of reference. Let them meet the plane at the points $O, T, U, V, W, X, Y, I$. Let $OT, OU, UV, TV, X, WY, YI, IX$ be joined. Then from the first part of this proposition, the parallelepiped $KV$ equals the parallelepiped $QI$. But from {{EuclidPropLink|book = XI|prop = 30|title = Parallelepipeds on Same Base and Same Height whose Extremities are not on Same Lines are Equal in Volume}}: :the parallelepiped $KV$ equals the parallelepiped $AE$ and: :the parallelepiped $QI$ equals the parallelepiped $CF$. Therefore the parallelepiped $AE$ is equal to the parallelepiped $CF$. {{qed}} {{Euclid Note|31|XI}}