Proof: By Euclid

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.
∎

Thank you to the contributors under CC BY-SA 4.0!

Github:
non-Github:: @Fitzpatrick

References

Adapted from (subject to copyright, with kind permission)

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

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

Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)