---
book: 11
number: 31
id: "XI.31"
kind: "theorem"
uses: ["[[book-1/proposition-23]]", "[[book-11/proposition-24]]", "[[book-11/proposition-29]]", "[[book-1/proposition-35]]", "[[book-5/proposition-7]]", "[[book-11/proposition-25]]", "[[book-5/proposition-11]]", "[[book-5/proposition-9]]", "[[book-11/proposition-30]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.31"
license: "CC-BY-SA-4.0"
---

# XI.31

*Parallelepipedal solids which are on equal bases and of the same height are equal to one another.*

## Proof

Let the parallelepipedal solids *AE*, *CF*, of the same height, be on equal bases *AB*, *CD*.

I say that the solid *AE* is equal to the solid *CF*.

First, let the sides which stand up, *HK*, *BE*, *AG*, *LM*, *PQ*, *DF*, *CO*, *RS*, be at right angles to the bases *AB*, *CD*; let the straight line *RT* be produced in a straight line with *CR*; on the straight line *RT*, and at the point *R* on it, let the angle *TRU* be constructed equal to the angle *ALB*, [[book-1/proposition-23|I. 23]] let *RT* be made equal to *AL*, and *RU* equal to *LB*, and let the base *RW* and the solid *XU* be completed.

Now, since the two sides *TR*, *RU* are equal to the two sides *AL*, *LB*, and they contain equal angles, therefore the parallelogram *RW* is equal and similar to the parallelogram *HL*.

Since again *AL* is equal to *RT*, and *LM* to *RS*, and they contain right angles, therefore the parallelogram *RX* is equal and similar to the parallelogram *AM*.

For the same reason *LE* is also equal and similar to *SU*; therefore three parallelograms of the solid *AE* are equal and similar to three parallelograms of the solid *XU*.

But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [[book-11/proposition-24|XI. 24]] therefore the whole parallelepipedal solid *AE* is equal to the whole parallelepipedal solid *XU*. [[book-11/definitions#Definition 10|XI. Def. 10]]

Let *DR*, *WU* be drawn through and meet one another at *Y*, let *aTb* be drawn through *T* parallel to *DY*, let *PD* be produced to *a*, and let the solids *YX*, *RI* be completed.

Then the solid *XY*, of which the parallelogram *RX* is the base and *Yc* its opposite, is equal to the solid *XU* of which the parallelogram *RX* is the base and *UV* its opposite, for they are on the same base *RX* and of the same height, and the extremities of their sides which stand up, namely *RY*, *RU*, *Tb*, *TW*, *Se*, *Sd*, *Xc*, *XV*, are on the same straight lines *YW*, *eV*. [[book-11/proposition-29|XI. 29]]

But the solid *XU* is equal to *AE*: therefore the solid *XY* is also equal to the solid *AE*.

And, since the parallelogram *RUWT* is equal to the parallelogram *YT* for they are on the same base *RT* and in the same parallels *RT*, *YW*, [[book-1/proposition-35|I. 35]] while *RUWT* is equal to *CD*, since it is also equal to *AB*, therefore the parallelogram *YT* is also equal to *CD*.

But *DT* is another parallelogram; therefore, as the base *CD* is to *DT*, so is *YT* to *DT*. [[book-5/proposition-7|V. 7]]

And, since the parallelepipedal solid *CI* has been cut by the plane *RF* which is parallel to opposite planes, as the base *CD* is to the base *DT*, so is the solid *CF* to the solid *RI*. [[book-11/proposition-25|XI. 25]]

For the same reason, since the parallelepipedal solid *YI* has been cut by the plane *RX* which is parallel to opposite planes, as the base *YT* is to the base *TD*, so is the solid *YX* to the solid *RI*. [[book-11/proposition-25|XI. 25]]

But, as the base *CD* is to *DT*, so is *YT* to *DT*; therefore also, as the solid *CF* is to the solid *RI*, so is the solid *YX* to *RI*. [[book-5/proposition-11|V. 11]]

Therefore each of the solids *CF*, *YX* has to *RI* the same ratio; therefore the solid *CF* is equal to the solid *YX*. [[book-5/proposition-9|V. 9]]

But *YX* was proved equal to *AE*; therefore *AE* is also equal to *CF*.

Next, let the sides standing up, *AG*, *HK*, *BE*, *LM*, *CN*, *PQ*, *DF*, *RS*, not be at right angles to the bases *AB*, *CD*; I say again that the solid *AE* is equal to the solid *CF*.

For from the points *K*, *E*, *G*, *M*, *Q*, *F*, *N*, *S* let *KO*, *ET*, *GU*, *MV*, *QW*, *FX*, *NY*, *SI* be drawn perpendicular to the plane of reference, and let them meet the plane at the points *O*, *T*, *U*, *V*, *W*, *X*, *Y*, *I*, and let *OT*, *OU*, *UV*, *TV*, *WX*, *WY*, *YI*, *IX* be joined.

Then the solid *KV* is equal to the solid *QI*, for they are on the equal bases *KM*, *QS* and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.]

But the solid *KV* is equal to the solid *AE*, and *QI* to *CF*; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [[book-11/proposition-30|XI. 30]]

Therefore the solid *AE* is also equal to the solid *CF*.

Therefore etc. Q. E. D.