hcju/proofwikiqa huggingface dataset:
:600px Let $AB$ and $CD$ be similar parallelepipeds. Let $AE$ be the side of $AB$ corresponding to the side $CF$ of $CD$. It is to be demonstrated that the parallelepiped $AB$ has to the parallelepiped $CD$ the ratio triplicate of that which $AE$ has to $CF$. That is: :$AB : CD = AE^3 : CF^3$ Let $EK, EL, EM$ be produced in a straight line with $AE, GE, HE$. Let $EK = CF, EL = FN, EM = FR$. Let the parallelogram $KL$ and parallelepiped $KP$ be completed. We have that the two sides $KE$ and $EL$ equal the two sides $CF$ and $FN$. Also $\angle KEL = \angle CFN$ because of similarity of $AB$ and $CD$. Therefore the parallelogram $KL$ is equal and similar to the parallelogram $CN$. For the same reason: :the parallelogram $KM$ is equal and similar to the parallelogram $CR$ and: :the parallelogram $EP$ is equal and similar to the parallelogram $DF$. Therefore from: :{{EuclidPropLink|book = XI|prop = 24|title = Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms}}: and: :{{EuclidDefLink|XI|10|Similar Equal Solid Figures}}: the whole parallelepiped $KP$ is equal and similar to the parallelepiped $CD$. Let the parallelogram $GK$ be completed. On the parallelograms $GK$ and $KL$ as bases, let the parallelepipeds $EO$ and $LQ$ be constructed with the same height as $AB$. We have that $AB$ and $CD$ are similar. So: :$AE : CF = EG : FN = EH : FR$ and: :$CF = EK$ :$FN : EL$ :$FR : EM$ Therefore: :$AE : EK = GE : EL = HE : EM$ But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$AE : EK = AG : GK$ :$GE : EL = GK : KL$ :$HE : EM = QE : KM$ Therefore: :$AG : GK = GK : KL = QE : LM$ But from {{EuclidPropLink|book = XI|prop = 32|title = Parallelepipeds of Same Height have Volume Proportional to Bases}}: :$AG : GK = AB : EO$ :$GK : KL = OE : QL$ :$QE : KM = QL : KP$ Therefore: :$AB : EO = EO : QL = QL : KP$ So from {{EuclidDefLink|V|10|Triplicate Ratio}}: :the parallelepipeds $AB$ has to the parallelepipeds $KP$ the ratio triplicate of that which $AE$ has to $EO$. But from {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$AB : EO = AG : GK = AE : EK$ Hence: :$AB : KP = AE^3 : EK^3$ But $KP = CD$ and $EK = CF$. Therefore the parallelepiped $AB$ has to the parallelepiped $CD$ the ratio triplicate of that which $AE$ has to $CF$. {{qed}} {{Euclid Note|33|XI}}
:500px Let the angles $\angle BAC$ and $\angle EDF$ be equal. From the points $A$ and $D$ let the elevated straight lines $AG$ and $DM$ be set up so they contain equal angles with the original straight lines: :$\angle MDE = \angle GAB$ :$\angle MDF = \angle GAC$ Let $G$ and $M$ be arbitrary points on $AG$ and $DM$. Let $GL$ and $MN$ be drawn from $G$ and $M$ perpendicular to the planes holding $\angle BAC$ and $\angle EDF$ respectively, meeting them at $L$ and $N$. Let $LA$ and $ND$ be joined. It is to be demonstrated that $\angle GAL = \angle MDN$. Let $AH = DM$. Let $HK$ be drawn through $H$ parallel to $GL$. But $GL$ is perpendicular to the plane holding $\angle BAC$. Therefore from {{EuclidPropLink|book = XI|prop = 8|title = Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane}}: :$HK$ is perpendicular to the plane holding $\angle BAC$. From the points $K, N$ let $KN, NF, KB, NE$ be drawn perpendicular to the straight lines $AC, DF, AB, DE$. Let $HC, CB, MF, FE$ be joined. From {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$HA^2 = HK^2 + KA^2$ and: :$KA^2 = KC^2 + CA^2$ Therefore: :$HA^2 = HK^2 + KC^2 + CA^2$ But from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$HC^2 = HK^2 + KC^2$ Therefore: :$HA^2 = HC^2 + CA^2$ Therefore from {{EuclidPropLink|book = I|prop = 48|title = Square equals Sum of Squares implies Right Triangle}}: :$\angle HCA$ is a right angle. For the same reason: :$\angle DFM$ is a right angle. Therefore: :$\angle HCA = \angle DFM$ But: :$\angle HAC = \angle MDF$ Therefore from {{EuclidPropLink|book = I|prop = 26|title = Triangle Side-Angle-Angle Equality}}: :$\triangle MDF = \triangle HAC$ Therefore: :$AC = DF$ Similarly it can be proved that $AB = DE$. So we have that: :$AC = DF$ :$AB = DE$ We also have that $\angle CAB = \angle FDE$ So from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\triangle CAB = \triangle FDE$ and so: :$\angle ACB = \angle DFE$ But the right angle $\angle ACK$ equals the right angle $\angle DFN$. Therefore the remaining angles are equal: :$\angle BCK = \angle EFN$ For the same reason: :$\angle CBK = \angle FEN$ Therefore from {{EuclidPropLink|book = I|prop = 26|title = Triangle Side-Angle-Angle Equality}}: :$\triangle BCK = \triangle EFN$ Therefore: :$CK = FN$ But we have that: :$AC = DF$ Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\triangle ACK = \triangle DFN$ and so: :$AK = DN$ We have that: :$AH = DM$ and so: :$AH^2 = DM^2$ But $\angle AKH$ is a right angle. So from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$AK^2 + KH^2 = AH^2$ Also $\angle DNM$ is a right angle. So from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$DM^2 + DN^2 = NM^2$ But as: :$AK = DN$ it follows that: :$AK^2 = DN^2$ Therefore: :$KH^2 = NM^2$ and so: :$KH = NM$ So we have that: :$AH = MD$ :$AK = DN$ :$HK = MN$ So from {{EuclidPropLink|book = I|prop = 8|title = Triangle Side-Side-Side Equality}}: :$\angle HAK = \angle MDN$ {{qed}} {{Euclid Note|35|XI}}
:500px Let $A, B, C$ be three straight lines in proportion: :$A : B  = B : C$ Let a solid angle be constructed at $E$ contained by the plane angles $\angle DEG, \angle GEF, \angle FED$. Let each of the straight lines $DE, GE, EF$ be made equal to $B$. Let the parallelepiped $EK$ be completed. Let the straight line $LM$ be made equal to $A$. Let a solid angle be constructed at $L$ contained by the plane angles $\angle NLO, \angle OLM, \angle MLN$. Let the straight line $LO$ be made equal to $B$. Let the straight line $LN$ be made equal to $C$. We have that: :$A : B = B : C$ while: :$A = LM$ :$B = LO = ED = EF$ :$C = LN$ Therefore: :$LM = EF = DE : LN$ Thus the sides about the equal angles $\angle NLM, \angle DEF$ are reciprocally proportional. Therefore from {{EuclidPropLink|book = VI|prop = 14|title = Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional}}: :the parallelogram $MN$ equals the parallelogram $DF$. It follows from {{EuclidPropLink|book = VI|prop = 35|qual=Porism|title = Condition for Equal Angles contained by Elevated Straight Lines from Plane Angles}}: :the perpendiculars from $G$ and $O$ to the planes containing $\angle NLM$ and $\angle DEF$ are equal. Therefore the parallelepipeds $LH$ and $EK$ are of equal height. So from {{EuclidPropLink|book = VI|prop = 31|title = Parallelepipeds on Equal Bases and Same Height are Equal in Volume}}: :$LH = EK$. We have that: :$LH$ is the parallelepiped formed out of $A, B, C$ :$EK$ is the parallelepiped formed out of $B$. Hence the result. {{qed}} {{Euclid Note|36|XI}}
:700px Let $AB$, $CD$, $EF$ and $GH$ be four straight lines in proportion, so that: :$AB : CD = EF : GH$ Let similar and similarly situated parallelepipeds $KA$, $LC$, $ME$ and $NG$ be described on $AB$, $CD$, $EF$ and $GH$ respectively. It is to be demonstrated that: :$KA : LC = ME : NG$ From {{EuclidPropLink|book = XI|prop = 33|title = Volumes of Similar Parallelepipeds are in Triplicate Ratio to Length of Corresponding Sides}}: :$KA : LC = AB^3 : CD^3$ For the same reason: :$ME : NG = EF^3 : GH^3$ We have that: :$AB : CD = EF : GH$ Therefore: :$KA : LC = ME : NG$ {{qed|lemma}} Suppose that: :$KA : LC = ME : NG$ Then from {{EuclidPropLink|book = XI|prop = 33|title = Volumes of Similar Parallelepipeds are in Triplicate Ratio to Length of Corresponding Sides}}: :$KA : LC = AB^3 : CD^3$ For the same reason: :$ME : NG = EF^3 : GH^3$ We have that: :$KA : LC = ME : NG$ Therefore: :$AB : CD = EF : GH$ {{qed}} {{Euclid Note|37|XI}}
:300px Let $AF$ be a cube. Let the edges of the opposite faces $CF$ and $AH$ of $AF$ be bisected at the points $K, L, M, N, O, P, Q, R$. Let the plane $KN$ be constructed through $K, L, M, N$. Let the plane $OR$ be constructed through $O, P, Q, R$. Let $US$ be the common section of $KN$ and $OR$. Let $DG$ be the diameter of the cube $AF$. It is to be demonstrated that: :$UT = TS$ :$DT = TG$ Let $DU, UE, BS, SG$ be joined. We have that $DO$ is parallel to $PE$. Therefore from {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Alternate Angles}}: :$\angle DOU = \angle UPE$ We have that: :$DO = PE$ :$OU = UP$ :$\angle DOU = \angle UPE$ So from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\triangle DOU = \triangle PUE$ Therefore: :$\angle OUD = \angle PUE$ So from {{EuclidPropLink|book = I|prop = 14|title = Two Angles making Two Right Angles make Straight Line}}: :$DUE$ is a straight line. {{explain|The statement above does not follow from the result stated -- this needs to be checked.}} For the same reason: :$BSG$ is a straight line. and: :$BS = SG$ We have that: :$CA = DB$ and $CA \parallel DB$ and: :$CA = EG$ and $CA \parallel EG$ Therefore from{{EuclidPropLink|book = I|prop = 9|title = Lines Parallel to Same Line not in Same Plane are Parallel to each other}}: :$DB = EG$ and $DB \parallel EG$ We have that the straight lines $DE$ and $BG$ join the endpoints of $DB$ and $EG$. Therefore from {{EuclidPropLink|book = I|prop = 33|title = Lines Joining Equal and Parallel Straight Lines are Parallel}}: :$DE \parallel BG$ Therefore from {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Alternate Angles}}: :$\angle EDT = \angle BGT$ Therefore from {{EuclidPropLink|book = I|prop = 29|title = Two Straight Lines make Equal Opposite Angles}}: :$\angle DTU = \angle GTS$ Therefore from {{EuclidPropLink|book = I|prop = 26|title = Triangle Side-Angle-Angle Equality}}: :$\triangle DTU = \triangle GTS$ Therefore: :$DT = TG$ and :$UT = TS$ {{qed}} {{Euclid Note|38|XI}}
:500px Let $ABCDEF$ and $GHKLMN$ be triangular prisms. Let $ABCDEF$ be positioned so that the parallelogram $AF$ can be identified as its base, so that its height is defined as the height of one of its triangular faces. Let $GHK$ be the triangular (conventionally defined) lower base of $GHKLMN$. Let the height of $ABCDEF$ equal the height of $GHKLMN$. Let the area of parallelogram $AF$ be twice the area of $GHK$. It is to be demonstrated that $ABCDEF$ and $GHKLMN$ are equal in volume. Let the parallelepipeds $AO$ and $GP$ be completed. We have that $AF$ is twice $\triangle GHK$. From {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :the parallelogram $HK$ is twice $\triangle GHK$. Therefore $AF = HK$. From {{EuclidPropLink|book = XI|prop = 31|title = Parallelepipeds on Equal Bases and Same Height are Equal in Volume}}: :the parallelepipeds $AO$ is equal to the parallelepipeds $GP$. But: :$ABCDEF$ is half of $AO$ and from {{EuclidPropLink|book = XI|prop = 28|title = Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected}}: :$GHKLMN$ is half of $GP$. Hence $ABCDEF$ is equal to $GHKLMN$. {{qed}} {{Euclid Note|39|XI|In the words of {{AuthorRef|Thomas Little Heath|Thomas L. Heath}}: :''The phraseology is interesting because we find one of the ''parallelogrammic'' faces of one of the triangular prisms called its ''base'', and the perpendicular on this plane from that vertex of either ''triangular'' face which is not in this plane its ''height''.}}
:600px Let $ABC$ and $FGH$ be circles. Let $ABCDE$ and $FGHKL$ be similar polygons inscribed in $ABC$ and $FGH$ respectively. Let $BM$ and $GN$ be diameters of $ABC$ and $FGH$ respectively. It is to be demonstrated that the ratio of the square on $BM$ to the square on $GN$ equals the ratio of the area of $ABCDE$ to the area of $FGHKL$. Let $BE, AM, GL, FN$ be joined. We have that $ABCDE$ is similar to $FGHKL$. Thus: :$\angle BAE = \angle GFL$ and so from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}: :$BA : AE = GF : FL$ Thus from {{EuclidPropLink|book = VI|prop = 6|title = Triangles with One Equal Angle and Two Sides Proportional are Similar}}: :$\triangle ABE$ is equiangular with $\triangle GFL$. Therefore: :$\angle AEB = \angle FLG$ But from {{EuclidPropLink|book = III|prop = 27|title = Angles on Equal Arcs are Equal}}: :$\angle AEB = \angle AMB$ and: :$\angle FLG = \angle FNG$ Therefore: :$\angle AMB = \angle FNG$ But from {{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}}: :the right angle $\angle BAM$ equals the right angle $\angle GFN$. Therefore from {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle ABM = \angle FGN$ Therefore $\triangle ABM$ is equiangular with $\triangle FGN$. Therefore from {{EuclidPropLink|book = VI|prop = 4|title = Equiangular Triangles are Similar}}: :$BM : GN = BA : GF$ But the ratio of the square on $BM$ to the square on $GN$ is the duplicate ratio of $BM$ to $GN$. From {{EuclidPropLink|book = VI|prop = 20|title = Similar Polygons are composed of Similar Triangles}}: :$ABCDE : FGHKL = BA^2 : GF^2$ Hence: :$ABCDE : FGHKL = BM^2 : GN^2$ Hence the result. {{qed}} {{Euclid Note|1|XII}}
:800px Let it be contrived that: :$S : ABCD = EFGH : T$ for some area $T$. It is to be demonstrated that $T$ is less than the circle $ABCD$. From the relationship: :$S : ABCD = EFGH : T$ it follows from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}} that: :$S : EFGH = ABCD : T$ But: :$S > EFGH$ Therefore: :$ABCD > T$ Hence the result. {{qed}} {{Euclid Note|2|XII}}
:700px Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$. It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of tetrahedron $ABCG$ to tetrahedron $DEFH$. Suppose the ratio of $ABCG$ to $DEFH$ is not equal to the ratio of $\triangle ABC$ to $\triangle DEF$. Then the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure $W$ which is either smaller than $DEFH$ or larger than $DEFH$. First suppose that $W$ is smaller than $DEFH$. From {{EuclidPropLink|book = XII|prop = 3|title = Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms}}: :let $DEFH$ be divided into two equal tetrahedra which are similar to $DEFH$ and two equal prisms. Then the two prisms are greater than half of $DEFH$. Again, let those tetrahedra be similarly divided. {{EuclidPropLink|book = X|prop = 1|title = Existence of Fraction of Number Smaller than Given}}: :Let this be done continually until there are left over from $DEFH$ some tetrahedra which are less than the excess by which $DEFH$ is greater than $W$. Let these, for the sake of argument, be $DQRS$ and $STUH$. Therefore their remainders, the prisms in $DEFH$, are greater than $W$. Let $ABCG$ be divided similarly and the same number of times as $DEFH$. From {{EuclidPropLink|book = XII|prop = 4|title = Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms}}: :the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prisms in $ABCG$ to the prisms in $DEFH$. But: :$\triangle ABC : \triangle DEF = ABCG : W$ So from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :the ratio of $ABCG$ to $W$ equals the ratio of the prisms in $ABCG$ to the prisms in $DEFH$. So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :the ratio of $ABCG$ to the prisms in it is equal to the ratio of $W$ to the prisms in $DEFH$. But $ABCG$ is greater than the prisms in it. Therefore $W > DEFG$. But by hypothesis $W < DEFG$. Therefore it is not the case that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure smaller than $DEFH$. Now suppose that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure $W$ greater than $DEFH$. Thus: :$\triangle DEF : \triangle ABC = W : ABCG$ But using the same technique as {{EuclidPropLink|book = XII|prop = 2|qual=Lemma|title = Areas of Circles are as Squares on Diameters}}: :$W : ABCG = DEFH : X$ where $X$ is some solid figure less than $ABCG$. Therefore from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$\triangle DEF : \triangle ABC = DEFG : X$ But this has been demonstrated to be absurd. Therefore it is not the case that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of $ABCG$ to some solid figure greater than $DEFH$. Hence the result. {{qed}} {{Euclid Note|5|XII}}
:600px Let there be two pyramids of the same height whose bases are the polygons $ABCDE$ and $FGHKL$ and whose apices are $M$ and $N$. It is to be demonstrated that the ratio of $ABCDE$ to $FGHKL$ equals the ratio of pyramid $ABCDEM$ to pyramid $FGHKLN$. Let $AC, AD, FH, FK$ be joined. We have that $ABCM$ and $ACDM$ are tetrahedra of the same height. So by {{EuclidPropLink|book = XII|prop = 5|title = Sizes of Tetrahedra of Same Height are as Bases}}: :the ratio of $ABCM$ to $ACDM$ equals the ratio of $\triangle ABC$ to $\triangle ACD$. From {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$ABCD : \triangle ACD = ABCDM : ACDM$ But by {{EuclidPropLink|book = XII|prop = 5|title = Sizes of Tetrahedra of Same Height are as Bases}}: :the ratio of $ACDM$ to $ADEM$ equals the ratio of $\triangle ACD$ to $\triangle ADE$. Therefore from {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$ABCD : \triangle ADE = ABCDM : ADEM$ Again from {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$ABCDE : \triangle ADE = ABCDEM : ADEM$ Similarly it can be shown that: :$FGHKL : \triangle FGH = FGHKLN : FGHN$ We have that $ADEM$ and $FGHN$ are two tetrahedra of the same height. Therefore from  {{EuclidPropLink|book = XII|prop = 5|title = Sizes of Tetrahedra of Same Height are as Bases}}: :the ratio of $ADEM$ to $FGHN$ equals the ratio of $\triangle ADE$ to $\triangle FGH$. But we have: :$\triangle ADE : ABCDE = ADEM : BCDEM$ Therefore by {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$ABCDE : \triangle FGH = ABCDEM : FGHN$ We also have: :$\triangle FGH : FGHKL = FGHN : FGHKLN$ Further by {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$ABCDE : FGHKL = ABCDEM : FGHKLN$ {{qed}} {{Euclid Note|6|XII}}
:300px Let $ABCDEF$ be a prism whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$. Then it is to be demonstrated that $ABCDEF$ can be divided into three equal tetrahedra. Let $BD, EC, CD$ be joined. We have that $ABED$ is a parallelogram. Therefore $BD$ is the diameter of $ABED$. So from {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :$\triangle ABD = \triangle EBD$ Therefore from {{EuclidPropLink|book = XII|prop = 5|title = Sizes of Tetrahedra of Same Height are as Bases}}: :the tetrahedron $ABDC$ whose base is $\triangle ABD$ and whose apex is $C$ equals the tetrahedron $DEBC$ whose base is $\triangle DEB$ and whose apex is $C$. But the tetrahedron $DEBC$ is the same as the tetrahedron $EBCD$ whose base is $\triangle EBC$ and whose apex is $D$. Therefore the tetrahedron $ABDC$ equals the tetrahedron $EBCD$. We have that $FCBE$ is a parallelogram. Therefore $CE$ is the diameter of $FCBE$. So from {{EuclidPropLink|book = I|prop = 34|title = Opposite Sides and Angles of Parallelogram are Equal}}: :$\triangle CEF = \triangle CBE$ Therefore from {{EuclidPropLink|book = XII|prop = 5|title = Sizes of Tetrahedra of Same Height are as Bases}}: :the tetrahedron $BCED$ whose base is $\triangle BCE$ and whose apex is $D$ equals the tetrahedron $ECFD$ whose base is $\triangle ECF$ and whose apex is $D$. But the tetrahedron $BCED$ was proved equal to the tetrahedron $ABDC$ whose base is $\triangle ABD$ and whose apex is $C$. Therefore the tetrahedron $CEFD$ whose base is $\triangle CEF$ and whose apex is $D$ equals the tetrahedron $ABDC$. Therefore the prism $ABCDEF$ has been divided into three equal tetrahedra. We have that the tetrahedron $ABDC$ is the same as the tetrahedron $CABD$ whose base is $\triangle CAB$ and whose apex is $D$. We have that the tetrahedron $ABDC$ is a third of the prism $ABCDEF$ whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$. Therefore tetrahedron $ABCD$ whose base is $\triangle ABC$ and whose apex is $D$ is a third of the prism $ABCDEF$ whose base is $\triangle ABC$ and whose opposite is $\triangle DEF$. {{qed}}
:500px Let $ABCG$ be a tetrahedron whose base is $\triangle ABC$ and whose apex is $G$. Let $DEFH$ be a tetrahedron whose base is $\triangle DEF$ and whose apex is $H$. Let $ABCG$ and $DEFH$ be similar and similarly situated. It is to be demonstrated that the ratio of $ABCG$ to $DEFH$ is triplicate of the ratio of the sides $BC$ and $EF$. Let the parallelepipeds $BGML$ and $EHQP$ be completed. We have that $ABCG$ is similar to $DEFH$. Therefore: :$\angle ABC = \angle DEF$ :$\angle GBC = \angle HEF$ :$\angle ABG = \angle DEH$ and: :$AB : DE = BC : EF = BG : EH$ It follows that the parallelogram $BM$ is similar to the parallelogram $EQ$. For the same reason: :the parallelogram $BN$ is similar to the parallelogram $ER$ and: :the parallelogram $BK$ is similar to the parallelogram $EO$ We also have that: :$BM, BK, BN$ are similar to $EQ, EO, ER$. But by {{EuclidPropLink|book = XI|prop = 24|title = Opposite Planes of Solid contained by Parallel Planes are Equal Parallelograms}}: :the three parallelograms $BM, BK, BN$ are equal and similar to their opposites and :the three parallelograms $EQ, EO, ER$ are equal and similar to their opposites. Therefore the parallelepipeds $BGML$ and $EHQP$ are contained by equal numbers of similar planes. Therefore $BGML$ is similar to $EHQP$. So by {{EuclidPropLink|book = XI|prop = 33|title = Volumes of Similar Parallelepipeds are in Triplicate Ratio to Length of Corresponding Sides}}: :$BGML : EQHP = BC^3 : EF^3$ But from: :{{EuclidPropLink|book = XI|prop = 28|title = Parallelepiped cut by Plane through Diagonals of Opposite Planes is Bisected}} and :{{EuclidPropLink|book = XII|prop = 7|title = Prism on Triangular Base divided into Three Equal Tetrahedra}} it follows that: :$BGML : EQHP = ABCG : DEFH$ Hence the result. {{qed}} {{Euclid Note|8|XII}}