In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal.

Let *AB* and *CD* be equal parallelepipedal solids.

I say that in the parallelepipedal solids *AB* and *CD* the bases are reciprocally proportional to the heights, that is the base *EH* is to the base *NQ* as the height of the solid *CD* is to the height of the solid *AB.*

First, let the sides which stand up, namely *AG, EF, LB, HK, CM, NO, PD,* and *QR,* be at right angles to their bases.

I say that the base *EH* is to the base *NQ* as *CM* is to *AG.*

If now the base *EH* equals the base *NQ,* and the solid *AB* equals the solid *CD,* then *CM* equal *AG.* For parallelepipedal solids of the same height are to one another as the bases, and the base *EH* is to *NQ* as *CM* is to *AG,* and it is clear that in the parallelepipedal solids *AB* and *CD* the bases are reciprocally proportional to the heights.

Next, let the base *EH* not be equal to the base *NQ,* but let *EH* be greater.

Now the solid *AB* equals the solid *CD,* therefore *CM* is also greater than *AG.*

Make *CT* equal to *AG,* complete the parallelepipedal solid *VC* on *NQ* as base with *CT* as height.

Now, since the solid *AB* equals the solid *CD,* and *CV* is outside them, and equals have to the same the same ratio, therefore the solid *AB* is to the solid *CV* as the solid *CD* is to the solid *CV.*

But the solid *AB* is to the solid *CY* as the base *EH* is to the base *NQ,* for the solids *AB* and *CV* are of equal height, and the solid *CD* is to the solid *CV* as the base *MQ* is to the base *TQ* and *CM* is to *CT,* therefore the base *EH* is to the base *NQ* as *MC* is to *CT.*

But *CT* equals *AG,* therefore the base *EH* is to the base *NQ* as *MC* is to *AG.*

Therefore in the parallelepipedal solids *AB* and *CD* the bases are reciprocally proportional to the heights.

Again, in the parallelepipedal solids *AB* and *CD* let the bases be reciprocally proportional to the heights, that is the base *EH* is to the base *NQ,* so let the height of the solid *CD* be to the height of the solid *AB.*

I say that the solid *AB* equals the solid *CD.*

Let the sides which stand up be at right angles to the bases.

Now, if the base *EH* equals the base *NQ,* and the base *EH* is to the base *NQ* as the height of the solid *CD* is to the height of the solid *AB,* therefore the height of the solid *CD* also equals the height of the solid *AB.*

But parallelepipedal solids on equal bases and of the same height equal one another, therefore the solid *AB* equals the solid *CD.*

Next, let the base *EH* not be equal to the base *NQ,* but let *EH* be greater.

Therefore the height of the solid *CD* is also greater than the height of the solid *AB,* that is, *CM* is greater than *AG.*

Make *CT* equal to *AG* again, and complete the solid *CV.*

Since the base *EH* is to the base *NQ* as *MC* is to *AG,* and *AG* equals *CT,* therefore the base *EH* is to the base *NQ* as *CM* is to *CT.*

But the base *EH* is to the base *NQ* as the solid *AB* is to the solid *CV,* for the solids *AB* and *CV* are of equal height, and *CM* is to *CT* as the base *MQ* is to the base *QT* and as the solid *CD* is to the solid *CV.*

Therefore the solid *AB* is to the solid *CV* as the solid *CD* to the solid *CV.* Therefore each of the solids *AB* and *CD* has to *CV* the same ratio. Therefore the solid *AB* equals the solid *CD.*

Now let the sides which stand up, *FE, BL, GA, HK, ON, DP, MC,* and *RQ,* not be at right angles to their bases. Draw perpendiculars from the points *F, G, B, K, O, M, D,* and *R* to the planes through *EH* and *NQ,* and let them meet the planes at *S, T, U, V, W, X, Y,* and *a.* Complete the solids *FV* and *Oa.*

I say that, in this case too, if the solids *AB* and *CD* are equal, then the bases are reciprocally proportional to the heights, that is, the base *EH* is to the base *NQ* as the height of the solid *CD* to the height of the solid *AB.*

Since the solid *AB* equals the solid *CD,* and *AB* equals *BT,* for they are on the same base *FK* and of the same height, and the solid *CD* equals *DX,* for they are again on the same base *RO* and of the same height, therefore the solid *BT* also equals the solid *DX.*

Above

Therefore the base *FK* is to the base *OR* as the height of the solid *DX* is to the height of the solid *BT.* But the base *FK* equals the base *EH,* and the base *OR* equals the base *NQ,* therefore the base *EH* is to the base *NQ* as the height of the solid *DX* is to the height of the solid *BT.*

But the solids *DX* and *BT* and the solids *DC* and *BA* have the same heights respectively, therefore the base *EH* is to the base *NQ* as the height of the solid *DC* is to the height of the solid *AB.*

Therefore in the parallelepipedal solids *AB* and *CD* the bases are reciprocally proportional to the heights.

Next, in the parallelepipedal solids *AB* and *CD* let the bases be reciprocally proportional to the heights, that is, as the base *EH* is to the base *NQ,* so let the height of the solid *CD* be to the height of the solid *AB.*

I say that the solid *AB* equals the solid *CD.*

With the same construction, since the base *EH* is to the base *NQ* as the height of the solid *CD* is to the height of the solid *AB,* and the base *EH* equals the base *FK,* and *NQ* equals *OR,* therefore the base *FK* is to the base *OR* as the height of the solid *CD* is to the height of the solid *AB.*

But the solids *AB* and *CD* and the solids *BT* and *DX* have the same heights respectively, therefore the base *FK* is to the base *OR* as the height of the solid *DX* is to the height of the solid *BT.*

Above

Therefore in the parallelepipedal solids *BT* and *DX* the bases are reciprocally proportional to the heights. Therefore the solid *BT* equals the solid *DX.*

But *BT* equals *BA,* for they are on the same base *FK* and of the same height, and the solid *DX* equals the solid *DC.* Therefore the solid *AB* also equals the solid *CD.*

Therefore, *in equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal.*

Q.E.D.

This proposition is used in the proof of proposition XII.9, an analogous statement about pyramids.