# Proof: By Euclid

• Let the solid angle at $E$, contained by $DEG$, $GEF$, and $FED$, be set out.
• And let $DE$, $GE$, and $EF$ each be made equal to $B$.
• And let the parallelepipedal solid $EK$ have been completed.
• And (let) $LM$ (be made) equal to $A$.
• And let the solid angle contained by $NLO$, $OLM$, and $MLN$ have been constructed on the straight line $LM$, and at the point $L$ on it, (so as to be) equal to the solid angle $E$ [Prop. 11.23].
• And let $LO$ be made equal to $B$, and $LN$ equal to $C$.
• And since as $A$ (is) to $B$, so $B$ (is) to $C$, and $A$ (is) equal to $LM$, and $B$ to each of $LO$ and $ED$, and $C$ to $LN$, thus as $LM$ (is) to $EF$, so $DE$ (is) to $LN$.
• And (so) the sides around the equal angles $NLM$ and $DEF$ are reciprocally proportional.
• Thus, parallelogram $MN$ is equal to parallelogram $DF$ [Prop. 6.14].
• And since the two plane rectilinear angles $DEF$ and $NLM$ are equal, and the raised straight lines stood on them (at their apexes), $LO$ and $EG$, are equal to one another, and contain equal angles respectively with the original straight lines (forming the angles), the perpendiculars drawn from points $G$ and $O$ to the planes through $NLM$ and $DEF$ (respectively) are thus equal to one another [Prop. 11.35 corr.] .
• Thus, the solids $LH$ and $EK$ (have) the same height.
• And parallelepiped solids on equal bases, and with the same height, are equal to one another [Prop. 11.31].
• Thus, solid $HL$ is equal to solid $EK$.
• And $LH$ is the solid (formed) from $A$, $B$, and $C$, and $EK$ the solid on $B$.
• Thus, the parallelepipedal solid (formed) from $A$, $B$, and $C$ is equal to the equilateral solid on $B$ (which is) equiangular with the aforementioned (solid).
• (Which is) the very thing it was required to show.

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