--- book: 11 number: 34 id: "XI.34" kind: "theorem" uses: ["[[book-11/proposition-32]]", "[[book-5/proposition-7]]", "[[book-11/proposition-25]]", "[[book-6/proposition-1]]", "[[book-11/proposition-31]]", "[[book-5/proposition-9]]", "[[book-11/proposition-29]]", "[[book-11/proposition-30]]"] source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.34" license: "CC-BY-SA-4.0" --- # XI.34 *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.* ## Proof Let *AB*, *CD* be equal parallelepipedal solids; I say that in the parallelepipedal solids *AB*, *CD* the bases are reciprocally proportional to the heights, that is, as the base *EH* is to the base *NQ*, so is the height of the solid *CD* to the height of the solid *AB*. First, let the sides which stand up, namely *AG*, *EF*, *LB*, *HK*, *CM*, *NO*, *PD*, *QR*, be at right angles to their bases; I say that, as the base *EH* is to the base *NQ*, so is *CM* to *AG*. If now the base *EH* is equal to the base *NQ*, while the solid *AB* is also equal to the solid *CD*, *CM* will also be equal to *AG*. For parallelepipedal solids of the same height are to one another as the bases; [[book-11/proposition-32|XI. 32]] and, as the base *EH* is to *NQ*, so will *CM* be to *AG*, and it is manifest that in the parallelepipedal solids *AB*, *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* is equal to the solid *CD*; therefore *CM* is also greater than *AG*. Let then *CT* be made equal to *AG*, and let the parallelepipedal solid *VC* be completed on *NQ* as base and with *CT* as height. Now, since the solid *AB* is equal to the solid *CD*, and *CV* is outside them, while equals have to the same the same ratio, [[book-5/proposition-7|V. 7]] therefore, as the solid *AB* is to the solid *CV*, so is the solid *CD* to the solid *CV*. But, as the solid *AB* is to the solid *CV*, so is the base *EH* to the base *NQ*, for the solids *AB*, *CV* are of equal height; [[book-11/proposition-32|XI. 32]] and, as the solid *CD* is to the solid *CV*, so is the base *MQ* to the base *TQ* [[book-11/proposition-25|XI. 25]] and *CM* to *CT* [[book-6/proposition-1|VI. 1]]; therefore also, as the base *EH* is to the base *NQ*, so is *MC* to *CT*. But *CT* is equal to *AG*; therefore also, as the base *EH* is to the base *NQ*, so is *MC* to *AG*. Therefore in the parallelepipedal solids *AB*, *CD* the bases are reciprocally proportional to the heights. Again, in the parallelepipedal solids *AB*, *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* is equal to the solid *CD*. Let the sides which stand up be again at right angles to the bases. Now, if the base *EH* is equal to the base *NQ*, and, as the base *EH* is to the base *NQ*, so is the height of the solid *CD* to the height of the solid *AB*, therefore the height of the solid *CD* is also equal to the height of the solid *AB*. But parallelepipedal solids on equal bases and of the same height are equal to one another; [[book-11/proposition-31|XI. 31]] therefore the solid *AB* is equal to 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*. Let *CT* be again made equal to *AG*, and let the solid *CV* be similarly completed. Since, as the base *EH* is to the base *NQ*, so is *MC* to *AG*, while *AG* is equal to *CT*, therefore, as the base *EH* is to the base *NQ*, so is *CM* to *CT*. But, as the base *EH* is to the base *NQ*, so is the solid *AB* to the solid *CV*, for the solids *AB*, *CV* are of equal height; [[book-11/proposition-32|XI. 32]] and, as *CM* is to *CT*, so is the base *MQ* to the base *QT* [[book-6/proposition-1|VI. 1]] and the solid *CD* to the solid *CV*. [[book-11/proposition-25|XI. 25]] Therefore also, as the solid *AB* is to the solid *CV*, so is the solid *CD* to the solid *CV*; therefore each of the solids *AB*, *CD* has to *CV* the same ratio. Therefore the solid *AB* is equal to the solid *CD*. [[book-5/proposition-9|V. 9]] Now let the sides which stand up, *FE*, *BL*, *GA*, *HK*, *ON*, *DP*, *MC*, *RQ*, not be at right angles to their bases; let perpendiculars be drawn from the points *F*, *G*, *B*, *K*, *O*, *M*, *D*, *R* to the planes through *EH*, *NQ*, and let them meet the planes at *S*, *T*, *U*, *V*, *W*, *X*, *Y*, *a*, and let the solids *FV*, *Oa* be completed; I say that, in this case too, if the solids *AB*, *CD* are equal, the bases are reciprocally proportional to the heights, that is, as the base *EH* is to the base *NQ*, so is the height of the solid *CD* to the height of the solid *AB*. Since the solid *AB* is equal to the solid *CD*, while *AB* is equal to *BT*, for they are on the same base *FK* and of the same height; [[book-11/proposition-29|XI. 29, 30]] and the solid *CD* is equal to *DX*, for they are again on the same base *RO* and of the same height; [*id.*] therefore the solid *BT* is also equal to the solid *DX*. Therefore, as the base *FK* is to the base *OR*, so is the height of the solid *DX* to the height of the solid *BT*. [Part 1.] But the base *FK* is equal to the base *EH*, and the base *OR* to the base *NQ*; therefore, as the base *EH* is to the base *NQ*, so is the height of the solid *DX* to the height of the solid *BT*. But the solids *DX*, *BT* and the solids *DC*, *BA* have the same heights respectively; therefore, as the base *EH* is to the base *NQ*, so is the height of the solid *DC* to the height of the solid *AB*. Therefore in the parallelepipedal solids *AB*, *CD* the bases are reciprocally proportional to the heights. Again, in the parallelepipedal solids *AB*, *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* is equal to the solid *CD*. For, with the same construction, since, as the base *EH* is to the base *NQ*, so is the height of the solid *CD* to the height of the solid *AB*, while the base *EH* is equal to the base *FK*, and *NQ* to *OR*, therefore, as the base *FK* is to the base *OR*, so is the height of the solid *CD* to the height of the solid *AB*. But the solids *AB*, *CD* and *BT*, *DX* have the same heights respectively; therefore, as the base *FK* is to the base *OR*, so is the height of the solid *DX* to the height of the solid *BT*. Therefore in the parallelepipedal solids *BT*, *DX* the bases are reciprocally proportional to the heights; therefore the solid *BT* is equal to the solid *DX*. [Part 1.] But *BT* is equal to *BA*, for they are on the same base *FK* and of the same height; [[book-11/proposition-29|XI. 29, 30]] and the solid *DX* is equal to the solid *DC*. [*id.*] Therefore the solid *AB* is also equal to the solid *CD*. Q. E. D.