◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-12-proportional-stereometry / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 12.03: Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms)

• Let there be a pyramid whose base is triangle $ABC$, and (whose) apex (is) point $D$.

• I say that pyramid $ABCD$ is divided into two pyramids having triangular bases (which are) equal to one another, and similar to the whole, and into two equal prisms.

• And the (sum of the) two prisms is greater than half of the whole pyramid.

• For let $AB$, $BC$, $CA$, $AD$, $DB$, and $DC$ have been cut in half at points $E$, $F$, $G$, $H$, $K$, and $L$ (respectively).
• And let $HE$, $EG$, $GH$, $HK$, $KL$, $LH$, $KF$, and $FG$ have been joined.
• Since $AE$ is equal to $EB$, and $AH$ to $DH$, $EH$ is thus parallel to $DB$ [Prop. 6.2].
• So, for the same (reasons), $HK$ is also parallel to $AB$.
• Thus, $HEBK$ is a parallelogram.
• Thus, $HK$ is equal to $EB$ [Prop. 1.34].
• But, $EB$ is equal to $EA$.
• Thus, $AE$ is also equal to $HK$.
• And $AH$ is also equal to $HD$.
• So the two (straight lines) $EA$ and $AH$ are equal to the two (straight lines) $KH$ and $HD$, respectively.
• And angle $EAH$ (is) equal to angle $KHD$ [Prop. 1.29].
• Thus, base $EH$ is equal to base $KD$ [Prop. 1.4].
• Thus, triangle $AEH$ is equal and similar to triangle $HKD$ [Prop. 1.4].
• So, for the same (reasons), triangle $AHG$ is also equal and similar to triangle $HLD$.
• And since $EH$ and $HG$ are two straight lines joining one another (which are respectively) parallel to two straight lines joining one another, $KD$ and $DL$, not being in the same plane, they will contain equal angles [Prop. 11.10].
• Thus, angle $EHG$ is equal to angle $KDL$.
• And since the two straight lines $EH$ and $HG$ are equal to the two straight lines $KD$ and $DL$, respectively, and angle $EHG$ is equal to angle $KDL$, base $EG$ [is] thus equal to base $KL$ [Prop. 1.4].
• Thus, triangle $EHG$ is equal and similar to triangle $KDL$.
• So, for the same (reasons), triangle $AEG$ is also equal and similar to triangle $HKL$.
• Thus, the pyramid whose base is triangle $AEG$, and apex the point $H$, is equal and similar to the pyramid whose base is triangle $HKL$, and apex the point $D$ [Def. 11.10] .
• And since $HK$ has been drawn parallel to one of the sides, $AB$, of triangle $ADB$, triangle $ADB$ is equiangular to triangle $DHK$ [Prop. 1.29], and they have proportional sides.
• Thus, triangle $ADB$ is similar to triangle $DHK$ [Def. 6.1] .
• So, for the same (reasons), triangle $DBC$ is also similar to triangle $DKL$, and $ADC$ to $DLH$.
• And since two straight lines joining one another, $BA$ and $AC$, are parallel to two straight lines joining one another, $KH$ and $HL$, not in the same plane, they will contain equal angles [Prop. 11.10].
• Thus, angle $BAC$ is equal to (angle) $KHL$.
• And as $BA$ is to $AC$, so $KH$ (is) to $HL$.
• Thus, triangle $ABC$ is similar to triangle $HKL$ [Prop. 6.6].
• And, thus, the pyramid whose base is triangle $ABC$, and apex the point $D$, is similar to the pyramid whose base is triangle $HKL$, and apex the point $D$ [Def. 11.9] .
• But, the pyramid whose base [is] [triangle]bookofproofs$6432 $HKL$, and apex the point $D$, was shown (to be) similar to the pyramid whose base is triangle $AEG$, and apex the point $H$.
• Thus, each of the pyramids $AEGH$ and $HKLD$ is similar to the whole pyramid $ABCD$.
• And since $BF$ is equal to $FC$, parallelogram $EBFG$ is double triangle $GFC$ [Prop. 1.41].
• And since, if two prisms (have) equal heights, and the former has a parallelogram as a base, and the latter a triangle, and the parallelogram (is) double the triangle, then the prisms are equal [Prop. 11.39], the prism contained by the two triangles $BKF$ and $EHG$, and the three parallelograms $EBFG$, $EBKH$, and $HKFG$, is thus equal to the prism contained by the two triangles $GFC$ and $HKL$, and the three parallelograms $KFCL$, $LCGH$, and $HKFG$.
• And (it is) clear that each of the prisms whose base (is) parallelogram $EBFG$, and opposite (side) straight line $HK$, and whose base (is) triangle $GFC$, and opposite (plane) triangle $HKL$, is greater than each of the pyramids whose bases are triangles $AEG$ and $HKL$, and apexes the points $H$ and $D$ (respectively), inasmuch as, if we [also] join the straight lines $EF$ and $EK$ then the prism whose base (is) parallelogram $EBFG$, and opposite (side) straight line $HK$, is greater than the pyramid whose base (is) triangle $EBF$, and apex the point $K$.
• But the pyramid whose base (is) triangle $EBF$, and apex the point $K$, is equal to the pyramid whose base is triangle $AEG$, and apex point $H$.
• For they are contained by equal and similar planes.
• And, hence, the prism whose base (is) parallelogram $EBFG$, and opposite (side) straight line $HK$, is greater than the pyramid whose base (is) triangle $AEG$, and apex the point $H$.
• And the prism whose base is parallelogram $EBFG$, and opposite (side) straight line $HK$, (is) equal to the prism whose base (is) triangle $GFC$, and opposite (plane) triangle $HKL$.
• And the pyramid whose base (is) triangle $AEG$, and apex the point $H$, is equal to the pyramid whose base (is) triangle $HKL$, and apex the point $D$.
• Thus, the (sum of the) aforementioned two prisms is greater than the (sum of the) aforementioned two pyramids, whose bases (are) triangles $AEG$ and $HKL$, and apexes the points $H$ and $D$ (respectively).
• Thus, the whole pyramid, whose base (is) triangle $ABC$, and apex the point $D$, has been divided into two pyramids (which are) equal to one another [and similar to the whole], and into two equal prisms.
• And the (sum of the) two prisms is greater than half of the whole pyramid.
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"