Pyramid Volume Theorem

Step 1 (triangular case by dissection). Take a triangular prism $ABC.A_1B_1C_1$ with base area $B$ and height $h$, so its volume is $V = Bh$ by the Prism Volume Theorem. Cut it along the two diagonal planes $(A_1BC)$ and $(A_1BC_1)$ into three tetrahedra: $A_1.ABC$, $A_1.BCC_1$ and $A_1.BB_1C_1$. A short computation shows these three tetrahedra have equal volume: $A_1.BCC_1$ and $A_1.BB_1C_1$ share apex $A_1$ over bases $BCC_1$ and $BB_1C_1$, which are congruent triangles (halves of the same parallelogram $BCC_1B_1$), so those two tetrahedra have equal volume; and comparing $A_1.ABC$ with $A_1.BCC_1$ (viewed as pyramids with apex $B$ or via the same congruent-base argument along the prism) shows all three parts are equal. Hence each tetrahedron has volume $\frac{1}{3} \cdot V_{\text{prism}} = \frac{1}{3}Bh$. Since $A_1.ABC$ is a triangular pyramid with base $ABC$ (area $B$) and apex $A_1$ at height $h$ above it, this proves $V = \frac{1}{3}Bh$ for triangular pyramids.

Step 2 (Cavalieri for the shape of the base). Now compare a triangular pyramid and a pyramid with any other base, both of base area $B$ and height $h$, apexes aligned at the same height. At height $y$ above the base $(0 \le y \le h)$, a plane parallel to the base cuts a pyramid in a copy of the base scaled by the factor $\left(\frac{h-y}{h}\right)$ — this is a standard similarity fact about central projection from the apex. Scaling a plane figure by a linear factor $k$ scales its area by $k^2$, so the cross-sectional area at height $y$ is $B\left(\frac{h-y}{h}\right)^2$ for every pyramid of base area $B$ and height $h$, regardless of the shape of the base.

Since the two pyramids being compared have identical cross-sectional area at every height, Cavalieri's principle gives them equal volume. As Step 1 established $V = \frac{1}{3}Bh$ for the triangular pyramid, the same formula $V = \frac{1}{3}Bh$ holds for a pyramid over any polygonal base of area $B$ and height $h$.