---
book: 11
number: 25
id: "XI.25"
kind: "theorem"
uses: ["[[book-11/proposition-24]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.25"
license: "CC-BY-SA-4.0"
---

# XI.25

*If a parallelepipedal solid be cut by a plane which is parallel to the opposite planes*, *then*, *as the base is to the base*, *so will the solid be to the solid.*

## Proof

For let the parallelepipedal solid *ABCD* be cut by the plane *FG* which is parallel to the opposite planes *RA*, *DH*; I say that, as the base *AEFV* is to the base *EHCF*, so is the solid *ABFU* to the solid *EGCD*.

For let *AH* be produced in each direction, let any number of straight lines whatever, *AK*, *KL*, be made equal to *AE*, and any number whatever, *HM*, *MN*, equal to *EH*; and let the parallelograms *LP*, *KV*, *HW*, *MS* and the solids *LQ*, *KR*, *DM*, *MT* be completed.

Then, since the straight lines *LK*, *KA*, *AE* are equal to one another, the parallelograms *LP*, *KV*, *AF* are also equal to one another, *KO*, *KB*, *AG* are equal to one another, and further *LX*, *KQ*, *AR* are equal to one another, for they are opposite. [[book-11/proposition-24|XI. 24]]

For the same reason the parallelograms *EC*, *HW*, *MS* are also equal to one another, *HG*, *HI*, *IN* are equal to one another, and further *DH*, *MY*, *NT* are equal to one another.

Therefore in the solids *LQ*, *KR*, *AU* three planes are equal to three planes.

But the three planes are equal to the three opposite; therefore the three solids *LQ*, *KR*, *AU* are equal to one another.

For the same reason the three solids *ED*, *DM*, *MT* are also equal to one another.

Therefore, whatever multiple the base *LF* is of the base *AF*, the same multiple also is the solid *LU* of the solid *AU*.

For the same reason, whatever multiple the base *NF* is of the base *FH*, the same multiple also is the solid *NU* of the solid *HU*.

And, if the base *LF* is equal to the base *NF*, the solid *LU* is also equal to the solid *NU*; if the base *LF* exceeds the base *NF*, the solid *LU* also exceeds the solid *NU*; and, if one falls short, the other falls short.

Therefore, there being four magnitudes, the two bases *AF*, *FH*, and the two solids *AU*, *UH*, equimultiples have been taken of the base *AF* and the solid *AU*, namely the base *LF* and the solid *LU*, and equimultiples of the base *HF* and the solid *HU*, namely the base *NF* and the solid *NU*, and it has been proved that, if the base *LF* exceeds the base *FN*, the solid *LU* also exceeds the solid *NU*, if the bases are equal, the solids are equal, and if the base falls short, the solid falls short.

Therefore, as the base *AF* is to the base *FH*, so is the solid *AU* to the solid *UH*. [[book-5/definitions#Definition 5|V. Def. 5]] Q. E. D.