hcju/proofwikiqa huggingface dataset:
:300px Let the two straight lines $AB$ and $CD$ be cut by the parallel planes $CH$, $KL$ and $MN$ at the points $A, E, B$ and $C, F, D$. It needs to be demonstrated that: :$AE : EB = CF : FD$ Let $AC$, $BD$ and $AD$ be joined. Let $AD$ meet $KL$ at $O$. Let $ED$ and $OF$ be joined. We have that the parallel planes $KL$ and $MN$ are cut by the plane $EBDO$. So by {{EuclidPropLink|book = XI|prop = 16|title = Common Sections of Parallel Planes with other Plane are Parallel}}: :their common sections $EO$ and $BD$ are parallel. Similarly, the parallel planes $GH$ and $KL$ are cut by the plane $AOFC$. So by {{EuclidPropLink|book = XI|prop = 16|title = Common Sections of Parallel Planes with other Plane are Parallel}}: :their common sections $AC$ and $OF$ are parallel. We have that the straight line $EO$ is parallel to $BD$. But $BD$ is one of the sides of $\triangle ABD$. Therefore from {{EuclidPropLink|book = VI|prop = 2|title = Parallel Line in Triangle Cuts Sides Proportionally}}: :$AE : EB = AO : OD$ Similarly, the straight line $OF$ is parallel to $AC$. But $AC$ is one of the sides of $\triangle ADC$. Therefore from {{EuclidPropLink|book = VI|prop = 2|title = Parallel Line in Triangle Cuts Sides Proportionally}}: :$AO : OD = CF : FD$ From {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : EB = CF : FD$ {{qed}} {{Euclid Note|17|XI}}
:350px Let $AB$ be an arbitrary straight line which is perpendicular to the plane of reference. It is to be demonstrated that every plane holding $AB$ is perpendicular to the plane of reference. Let an arbitrary plane $DE$ be drawn through $AB$. Let $CE$ be the common section of the plane $DE$ and the plane of reference. Let $F$ be an arbitrary point on $CE$. From {{EuclidPropLink|book = I|prop = 11|title = Construction of Perpendicular Line}}: :let $FG$ be drawn in $DE$ perpendicular to $CE$. We have that $AB$ is perpendicular to the plane of reference. Therefore from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}:  :$AB$ is perpendicular to all the straight lines which meet it and are in the plane of reference. Therefore $AB$ is perpendicular to $CE$. Therefore $\angle ABF$ is a right angle. But $\angle GFB$ is also a right angle. Therefore by {{EuclidPropLink|book = I|prop = 28|title = Supplementary Interior Angles implies Parallel Lines}}: :$AB \parallel FG$ But $AB$ is perpendicular to the plane of reference. Therefore from {{EuclidPropLink|book = XI|prop = 8|title = Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane}}: : $FG$ is perpendicular to the plane of reference. Thus the conditions are fulfilled for {{EuclidDefLink|XI|4|Plane at Right Angles to Plane}} to apply: :$DE$ is perpendicular to the plane of reference. As $DE$ is arbitrary, the result follows. {{qed}} {{Euclid Note|18|XI}}
:400px Let $AB$ and $BC$ be planes which are perpendicular to the plane of reference. Let $BD$ be the common section of $AB$ and $BC$. It is to be demonstrated that $BD$ is perpendicular to the plane of reference. Suppose $BD$ is not perpendicular to the plane of reference. From the point $D$, let: :$DE$ be drawn in the plane $AB$ perpendicular to $AD$ :$DF$ be drawn in the plane $AC$ perpendicular to $CD$ We have that: :$AB$ is perpendicular to the plane of reference and: :$DE$ is in the plane $AB$ perpendicular to $AD$. Therefore from {{EuclidDefLink|XI|4|Plane at Right Angles to Plane}}: :$DE$ is perpendicular to the plane of reference. Similarly it can be proved that $DF$ is also perpendicular to the plane of reference. Therefore from $D$ we have two straight lines have been set up perpendicular to the plane of reference on the same side. This contradicts {{EuclidPropLink|book = XI|prop = 13|title = Straight Line Perpendicular to Plane from Point is Unique}}. Hence the result. {{qed}} {{Euclid Note|19|XI}}
:350px Let $A$ be a solid angle which is contained by the three plane angles $\angle BAC$, $\angle CAD$ and $\angle DAB$. It is to be demonstrated that any two of these plane angles together are greater than the third. Suppose $\angle BAC$, $\angle CAD$ and $\angle DAB$ are all equal. Then any two together are greater than the remaining one and the result follows. Suppose otherwise. {{WLOG}}, suppose $\angle BAC$ is greater than $\angle BAD$. Let $E$ be constructed on $AB$ such that $\angle BAE$, in the plane through $BA$ and $AC$, is equal to $\angle DAB$. Let $AE = AD$. Let $BEC$ be drawn across through $E$ such that the straight lines $AB$ and $AC$ are cut at $B$ and $C$. Let $DB$ and $DC$ be joined. We have that: :$DA = AE$ and $AB$ is common. So we have two sides equal to two sides, while $\angle DAB = \angle DAE$. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$DB = BE$ We have from {{EuclidPropLink|book = I|prop = 20|title = Sum of Two Sides of Triangle Greater than Third Side}}: :$BD + DC > BC$ But we have that $BD = BE$. Therefore $DC > EC$. We have that: :$DA = AE$ and $AC$ is common. We also have that $DC > DE$. Therefore from {{EuclidPropLink|book = I|prop = 25|title = Converse Hinge Theorem}}: :$\angle DAC > \angle EAC$ But: :$\angle DAB = \angle BAE$ and so: :$\angle DAB + \angle DAC > \angle BAC$ Similarly it can be shown that the remaining plane angles, taken two together, are greater than the remaining one. {{qed}} {{Euclid Note|20|XI}}
:300px Let $A$ be a solid angle which is contained by the three plane angles $\angle BAC$, $\angle CAD$ and $\angle DAB$. It is to be demonstrated that $\angle BAC + \angle CAD + \angle DAB$ is less than $4$ right angles. Let $B$, $C$ and $D$ be arbitrary points on the straight lines $AB$, $AC$ and $AD$. Let $BC$, $CD$ and $DB$ be joined. We now have that the solid angle at $B$ is contained by the three plane angles $\angle CBA$, $\angle ABD$ and $\angle CBD$. From {{EuclidPropLink|book = XI|prop = 20|title = Sum of Two Angles of Three containing Solid Angle is Greater than Other Angle}}: :any two of these is greater than the other one. Therefore $\angle CBA + \angle ABD > \angle CBD$. For the same reason: :$\angle BCA + \angle ACD > \angle BCD$ :$\angle CDA + \angle ADB > \angle CDB$ Therefore the sum of the six plane angles: :$\angle CBA + \angle ABD + \angle BCA + \angle ACD + \angle CDA + \angle ADB$ are greater than the three plane angles: :$\angle CBD + \angle BCD + \angle CDB$ But from {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle CBD + \angle BCD + \angle CDB$ equals $2$ right angles. Therefore $\angle CBA + \angle ABD + \angle BCA + \angle ACD + \angle CDA + \angle ADB$ is greater than two right angles. Also from {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle ABC + \angle ACD + \angle ADB$ equals $2$ right angles. Therefore from {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle CBA + \angle ACB + \angle BAC + \angle ACD + \angle CDA + \angle CAD + \angle ADB + \angle DBA + \angle BAD$ equals $6$ right angles. Of them, $\angle ABC + \angle BCA + \angle ACD + \angle CDA + \angle ADB + \angle DBA$ are greater than $2$ right angles. Therefore the remaining three angles: :$\angle BAC + \angle CAD + \angle DAB$ are less than $4$ right angles. {{qed}} {{Euclid Note|21|XI}}
:600px Let $\angle ABC, \angle DEF, \angle GHK$ be plane angles such that the sum of any two is greater than the remaining one. That is: :$\angle ABC + \angle DEF > \angle GHK$ :$\angle DEF + \angle GHK > \angle ABC$ :$\angle GHK + \angle ABC > \angle DEF$ Let the straight lines $AB, BC, DE, EF, GH, GK$ be equal. Let $AC$, $DF$ and $GK$ be joined. It is to be demonstrated that it is possible to construct a triangle from straight lines equal to $AC$, $DF$ and $HK$. If $\angle ABC, \angle DEF, \angle GHK$ are equal to one another, then it is possible to construct an equilateral triangle from $AC$, $DF$ and $HK$. :250pxright Otherwise, let $\angle ABC, \angle DEF, \angle GHK$ be unequal. On the straight line $HK$ at the point $H$, let: :$\angle KHL$ be constructed equal to $\angle ABC$ :$HL$ be constructed equal to one of $AB, BC, DE, EF, GH, GK$ :$KL$ be joined. We have that: :$AB$ and $BC$ are equal to $KH$ and $HL$ :$\angle ABC = \angle KHL$ Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$AC = KL$ We have that: :$\angle GHK + \angle ABC > \angle DEF$ while: :$\angle ABC = \angle KHL$ Therefore: :$\angle GHL > \angle DEF$ We have that: :$GH$ and $HL$ are equal to $DE$ and $EF$ :$\angle GHL > \angle DEF$ Therefore from {{EuclidPropLink|book = I|prop = 24|title = Hinge Theorem}}: :$GL > DF$ But: :$GK + KL > GL$ Therefore: :$GK + KL > DF$ But: :$KL = AC$ Therefore: :$AC + GK > DF$ Similarly it can be proved that: :$AC + DF > GK$ and that: :$DF + GK > AC$ Hence the result. {{qed}} {{Euclid Note|22|XI}}
:400px Let the straight lines $AB$ and $LO$ be set out. Le $AB > LO$. Let the semicircle $ACB$ be described on $AB$. :400px Using {{EuclidPropLink|book = IV|prop = 1|title = Fitting Chord Into Circle}}: :Let $AC = LO$ be fitted into the semicircle $ACB$. Let $BC$ be joined. From {{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}}: :$\angle ACB$ is a right angle. Therefore from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$AB^2 = AC^2 + CB^2$ Hence: :$AB^2 - AC^2 = CB^2$ But $AC = LO$. Therefore: :$AB^2 - LO^2 = CB^2$ So if we then cut of $OR = BC$: :$AB^2 - LO^2 = OR^2$ {{qed}} {{Euclid Note|23|XI}}
:450px Let the solid $CDHG$ be contained by the parallel planes $AC, GF, AH, DF, BF, AE$. It is to be demonstrated that opposite planes are equal parallelograms. We have that the two parallel planes $BG$ and $CE$ are cut by the plane $AC$. From {{EuclidPropLink|book = XI|prop = 16|title = Common Sections of Parallel Planes with other Plane are Parallel}}: :the common sections of $BG$ and $CE$ are parallel lines. Thus $AB \parallel DC$. Again, we have that the two parallel planes $BF$ and $AE$ are cut by the plane $AC$. Thus $BC \parallel AD$. But $AB \parallel DC$. Therefore $AC$ is by definition a parallelogram. Similarly it can be shown that each of $GF, AH, DF, BF, AE$ are parallelograms. Let $AH$ and $DF$ be joined. We have that: :$AB \parallel DC$ and: :$BH \parallel CF$ Thus the two straight lines $AB$ and $BH$ which meet one another are parallel to the two straight lines $DC$ and $CF$ which also meet one another, but not in the same plane. Therefore by {{EuclidPropLink|book = XI|prop = 10|title = Two Lines Meeting which are Parallel to Two Other Lines Meeting contain Equal Angles}}: :$\angle ABH = \angle DCF$ From {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :$AB$ and $BH$ are equal to $DC$ and $CF$. and because: :$\angle ABH = \angle DCF$ it follows that: :$AH = DF$ So from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\triangle ABH = \triangle DCF$ We have that the parallelogram $BG$ is double $\triangle ABH$. From {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :the parallelogram $CE$ is double the $\triangle DCF$. Therefore the parallelogram $BG$ equals the parallelogram $CE$. Similarly it is shown that: :the parallelogram $AC$ equals the parallelogram $GF$ and: :the parallelogram $AE$ equals the parallelogram $BF$. {{qed}} {{Euclid Note|24|XI}}
:600px Let the parallelepiped $ABCD$ be cut by the plane $FG$ which is parallel to the opposite planes $RA$ and $DH$. It is to be demonstrated that the ratio of the base $AEFV$ to the base $EHCF$ equals the ratio of the parallelepiped $ABFU$ to the parallelepiped $EGCD$. Let $AH$ be produced in each direction. Let any number of straight lines $AK, KL$ equal to $AE$ be made. Let any number of straight lines $HM, MN$ equal to $EH$ be made. Let the parallelograms $LP, KV, HW, MS$ be completed. Let the parallelepipeds $LQ, KR, DM, MT$ be completed. We have that the straight lines $LK, KA, AE$ are equal. Therefore the parallelograms $LP, KV, AF$ are equal. From {{EuclidPropLink|book = XI|prop = 24|title = Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms}}: :the parallelograms $LX, KQ, AR$ are equal. For the same reason, the parallelograms $EC, HW, MS$ are equal. Therefore the parallelepipeds $HG, HI, NT$ are equal. Further, the parallelograms $DH, MY, MT$ are equal. Therefore, in the parallelepipeds $LQ, KR, AU$, three planes equal three planes. But the three planes equal the three planes opposite. Therefore the three parallelepipeds $LQ, KR, AU$ are equal to one another. For the same reason, the three parallelepipeds $ED, DM, MT$ are equal to one another. Therefore, whatever multiple the base $LF$ is of the base $AF$, the same multiple also is the parallelepiped $LU$ of the parallelepiped $AU$. For the same reason, whatever multiple the base $NF$ is of the base $FH$, the same multiple also is the parallelepiped $NU$ of the parallelepiped $HU$. And if the base $LF$ equals the base $NF$, the parallelepiped $LU$ equals the parallelepiped $NU$. If the base $LF$ exceeds the base $NF$, the parallelepiped $LU$ exceeds the parallelepiped $NU$. If the base $LF$ falls short of the base $NF$, the parallelepiped $LU$ falls short of the parallelepiped $NU$. We have four magnitudes: :the two bases $AF$ and $FH$ and: :the two parallelepipeds $AU$ and $HU$. Equimultiples have been taken of the base $AF$ and the parallelepiped $AU$: the base $LF$ and the parallelepiped $LU$. and equimultiples have been taken of the base $HF$ and the parallelepiped $HU$: the base $NF$ and the parallelepiped $NU$. It has been proved that: : if the base $LF$ exceeds the base $FN$, the parallelepiped $LU$ exceeds the parallelepiped $NU$ :if the base $LF$ equals the base $NF$, the parallelepiped $LU$ equals the parallelepiped $NU$ :if the base $LF$ falls short of the base $NF$, the parallelepiped $LU$ falls short of the parallelepiped $NU$. Therefore, as the base $AF$ is to the base $FH$, so the parallelepiped $AU$ is to the parallelepiped $UH$. {{qed}} {{Euclid Note|25|XI}}
:400px Let $AB$ be the given straight line. Let $A$ be the given point on $AB$. Let the angle at $D$, contained by the plane angles $\angle EDC, \angle EDF$ and $\angle FDC$, be the given solid angle. Let an arbitrary point $F$ be taken on $DF$. Using {{EuclidPropLink|book = XI|prop = 11|title = Construction of Straight Line Perpendicular to Plane from point not on Plane}}: :Let $FG$ be drawn perpendicular to the plane through $ED$ and $DC$, meeting the plane at $G$. Let $DG$ be joined. Using {{EuclidPropLink|book = I|prop = 11|title = Construction of Equal Angle}}: :Let $\angle BAL$ be constructed on the straight line $AB$ and at the point $A$ equal to $\angle EDC$, and $\angle BAK$ equal to $\angle EDG$. Let $AK = DG$. Using {{EuclidPropLink|book = XI|prop = 12|title = Construction of Straight Line Perpendicular to Plane from point on Plane}}: :Let $KH$ be set up from the point $K$ perpendicular to the plane through $BA$ and $AL$. Let $KH = GF$. Let $HA$ be joined. It is to be demonstrated that the solid angle at $A$ contained by the plane angles $\angle BAL \angle BAH$ and $\angle HAL$ equals the angle at $D$, contained by the plane angles $\angle EDC, \angle EDF$ and $\angle FDC$. Let $AB$ and $DE$ be cut off equal to each other. Let $HB, KB, FE, GE$ be joined. We have that $FG$ is perpendicular to the plane of reference. So by {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$FG$ is perpendicular to all straight lines which meet it and are in the plane of reference. Therefore $\angle FGD$ and $\angle FGE$ are right angles. For the same reason, $\angle HKA$ and $\angle HKB$ are right angles. We have that $KA$ and $AB$ are equal to $GD$ and $DE$ respectively. From {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$KB = GE$ But also: :$KH = GF$ Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$HB = FE$ We have that $AK$ and $KH$ are equal to $DG$ and $GF$ respectively. From {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$AH = FD$ But also: :$AB = DE$ Therefore $HA$ and $AB$ are equal to $DF$ and $DE$. We also have that $HB = FE$. Therefore from {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle BAH = \angle EDF$ For the same reason: :$\angle HAL = \angle FDC$ and: :$\angle BAL = \angle EDC$ Hence the result. {{qed}} {{Euclid Note|26|XI}}
:500px Let $AB$ be the given straight line. Let $CD$ be the given parallelepiped. Using {{EuclidPropLink|book = XI|prop = 26|title = Construction of Solid Angle equal to Given Solid Angle}}: :On the straight line $AB$ let a solid angle contained by the plane angles $\angle BAH, \angle HAK, \angle KAB$ be constructed equal to the solid angle at $C$ contained by the plane angles $\angle ECF, \angle ECG, \angle GCF$ such that: ::$\angle BAH = \angle ECF$ ::$\angle KAH = \angle GCF$ ::$\angle BAK = \angle ECG$ Using {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}: :Let it be contrived that: ::$EC : CG = BA : AK$ and: ::$GC : CF = KA : AH$ Therefore from {{EuclidPropLink|book = V|prop = 12|title = Equality of Ratios Ex Aequali}}: :$EC : CF = BA : AH$ Let the parallelogram $HB$ and the parallelepiped $AL$ be completed. We have that: :$EC : CG = BA : AK$ and the sides about the equal angles $\angle BAK$ and $\angle ECG$ are proportional. Therefore the parallelogram $GE$ is similar to the parallelogram $KB$. For the same reason: :the parallelogram $KH$ is similar to the parallelogram $GF$. :the parallelogram $FE$ is similar to the parallelogram $HB$. Therefore three parallelograms of the parallelepiped $CD$ are similar to the three parallelograms of the parallelepiped $AL$. But the former three are equal and similar to the three opposite parallelograms. Therefore the whole parallelepiped $CD$ is similar to the whole parallelepiped $AL$. {{qed}} {{Euclid Note|27|XI}}
:300px Let the parallelepiped $AB$ be cut by the plane $CDEF$ through the diagonals $CF$ and $DE$ of opposite planes. It is to be demonstrated that the parallelepiped $AB$ is bisected by the plane $CDEF$ From {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :$\triangle CGF = \triangle CFB$ and: :$\triangle ADE = \triangle DEH$ while: :the parallelogram $CA$ is equal to the parallelogram $EB$ and: :the parallelogram $GE$ is equal to the parallelogram $CH$ as they are opposite. Therefore from {{EuclidDefLink|XI|10|Similar Equal Solid Figures}}: :the prism contained by $\triangle CGF$ and $\triangle ADE$ and the three parallelograms $GE, AC, AE$ equals: :the prism contained by $\triangle CFB$ and $\triangle DEH$ and the three parallelograms $CH, BE, CE$. Hence the result. {{qed}} {{Euclid Note|28|XI}}
:400px Let $CM$ and $CN$ be parallelepipeds on the same bases and of the same height. Let the endpoints of their vertical sides: :$AG, AF, LM, LN, CD, CE, BH, BK$ be on the same straight lines $FN$ and $DK$. It is to be demonstrated that the parallelepiped $CM$ is equal to the parallelepiped $CN$. We have that each of the figures $CH$ and $CK$ is a parallelogram. Thus from {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :$CB$ equals each of the straight lines $DH$ and $EK$. Thus $DH = EK$. Let $EH$ be subtracted from both $DH$ and $EK$. Therefore their remainders $DE$ and $HK$ are equal. So from: :{{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}} and: :{{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}} it follows that: :$\triangle DCE = \triangle HBK$ and from {{EuclidPropLink|book = I|prop = 36|title = Parallelograms with Equal Base and Same Height have Equal Area}}: :the parallelogram $DG$ equals the parallelogram $HN$. For the same reason: :$\triangle AFG = \triangle MLN$ But the parallelogram $CF$ equals the parallelogram $BM$, as they are opposite. Therefore: :the prism contained by $\triangle AFG$ and $\triangle DCE$ and the three parallelograms $AD, DG, CG$ equals: :the prism contained by $\triangle MLN$ and $\triangle HBK$ and the three parallelograms $BM, HN, BN$. Let there be added to each the solid of which the parallelogram $AB$ is its base and $GEHM$ its opposite. Therefore the whole parallelepiped $CM$ is equal to the whole parallelepiped $CN$. {{qed}} {{Euclid Note|29|XI}}
:400px Let $CM$ and $CN$ be parallelepipeds on the same base $AB$ and of the same height. Let the endpoints of their vertical sides: :$AF, AG, LM, LN, CD, CE, BH, BK$ be not on the same straight lines. It is to be demonstrated that the parallelepiped $CM$ is equal to the parallelepiped $CN$. Let $NK$ and $DH$ be produced, and meet each other at $R$. Let $FM$ and $GE$ be produced to $P$ and $Q$. Let $AO, LP, CQ, BR$ be joined. We have that $CM$ and $CP$ are on the same parallelogram $ABCL$ and of the same height, and the endpoints of their vertical sides: :$AF, AO, LM, LP, CD, CQ, BH, BR$ are on the same straight lines $FP$ and $FR$. So 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 $CM$, of which the parallelogram $ABCL$ is the base and $FDHM$ the opposite is equal to :the parallelepiped $CP$, of which the parallelogram $ABCL$ is the base and $OQRP$ the opposite. But we also have that $CP$ and $CN$ are on the same parallelogram $ABCL$ and of the same height, and the endpoints of their vertical sides: :$AG, AO, LN, LP, CE, CQ, BK, BR$ are on the same straight lines $GQ$ and $NR$. So 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 $CN$, of which the parallelogram $ABCL$ is the base and $ACBL$ the opposite is equal to :the parallelepiped $CP$, of which the parallelogram $ABCL$ is the base and $OQRP$ the opposite. Hence the parallelepiped $CM$ equals the parallelepiped $CN$. {{qed}} {{Euclid Note|30|XI}}
:500px Let $AB$ and $CD$ be parallelepipeds of the same height. Let $AE$ be the base of $AB$ and $CF$ be the base of $AB$. Using {{EuclidPropLink|book = I|prop = 45|title = Construction of Parallelogram in Given Angle equal to Given Polygon}}: :Let the parallelogram $FH$ equal in area to $AE$ be applied to $FG$. Let the parallelepiped $GK$ be constructed with the same height as $CD$ on the base $FH$. Then from {{EuclidPropLink|book = XI|prop = 31|title = Parallelepipeds on Equal Bases and Same Height are Equal in Volume}}: :the parallelepiped $AB$ equals the parallelepiped $GH$. We have that the parallelepiped $CK$ is cut by the plane $DG$ which is parallel to opposite planes. Therefore from {{EuclidPropLink|book = XI|prop = 25|title = Parallelepiped cut by Plane Parallel to Opposite Planes}}: :the parallelepiped $CD$ is to the parallelepiped $DH$ as the base $CF$ is to the base $FH$. But: :the base $FH$ equals the base $AE$ and: :the parallelepiped $GK$ equals the parallelepiped $AB$. Therefore, as the base $AE$ is to the base $CF$, the parallelepiped $AB$ is to the parallelepiped $CD$. {{qed}} {{Euclid Note|32|XI}}