---
book: 2
number: 9
id: "II.9"
kind: "theorem"
uses: ["[[book-1/proposition-32]]", "[[book-1/proposition-29]]", "[[book-1/proposition-6]]", "[[book-1/proposition-47]]", "[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:2.prop.9"
license: "CC-BY-SA-4.0"
---

# II.9

*If a straight line be cut into equal and unequal segments*, *the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section*.

## Proof

For let a straight line *AB* be cut into equal segments at *C*, and into unequal segments at *D*;

I say that the squares on *AD*, *DB* are double of the squares on *AC*, *CD*.

For let *CE* be drawn from *C* at right angles to *AB*, and let it be made equal to either *AC* or *CB*; let *EA*, *EB* be joined, let *DF* be drawn through *D* parallel to *EC*, and *FG* through *F* parallel to *AB*, and let *AF* be joined.

Then, since *AC* is equal to *CE*, the angle *EAC* is also equal to the angle *AEC*.

And, since the angle at *C* is right, the remaining angles *EAC*, *AEC* are equal to one right angle. [[book-1/proposition-32|I. 32]]

And they are equal; therefore each of the angles *CEA*, *CAE* is half a right angle.

For the same reason each of the angles *CEB*, *EBC* is also half a right angle; therefore the whole angle *AEB* is right.

And, since the angle *GEF* is half a right angle, and the angle *EGF* is right, for it is equal to the interior and opposite angle *ECB*, [[book-1/proposition-29|I. 29]] the remaining angle *EFG* is half a right angle; [[book-1/proposition-32|I. 32]] therefore the angle *GEF* is equal to the angle *EFG*, so that the side *EG* is also equal to *GF*. [[book-1/proposition-6|I. 6]]

Again, since the angle at *B* is half a right angle, and the angle *FDB* is right, for it is again equal to the interior and opposite angle *ECB*, [[book-1/proposition-29|I. 29]] the remaining angle *BFD* is half a right angle; [[book-1/proposition-32|I. 32]] therefore the angle at *B* is equal to the angle *DFB*, so that the side *FD* is also equal to the side *DB*. [[book-1/proposition-6|I. 6]]

Now, since *AC* is equal to *CE*, the square on *AC* is also equal to the square on *CE*; therefore the squares on *AC*, *CE* are double of the square on *AC*.

But the square on *EA* is equal to the squares on *AC*, *CE*, for the angle *ACE* is right; [[book-1/proposition-47|I. 47]] therefore the square on *EA* is double of the square on *AC*.

Again, since *EG* is equal to *GF*, the square on *EG* is also equal to the square on *GF*; therefore the squares on *EG*, *GF* are double of the square on *GF*.

But the square on *EF* is equal to the squares on *EG*, *GF*; therefore the square on *EF* is double of the square on *GF*.

But *GF* is equal to *CD*; [[book-1/proposition-34|I. 34]] therefore the square on *EF* is double of the square on *CD*.

But the square on *EA* is also double of the square on *AC*; therefore the squares on *AE*, *EF* are double of the squares on *AC*, *CD*.

And the square on *AF* is equal to the squares on *AE*, *EF*, for the angle *AEF* is right; [[book-1/proposition-47|I. 47]] therefore the square on *AF* is double of the squares on *AC*, *CD*.

But the squares on *AD*, *DF* are equal to the square on *AF*, for the angle at *D* is right; [[book-1/proposition-47|I. 47]] therefore the squares on *AD*, *DF* are double of the squares on *AC*, *CD*.

And *DF* is equal to *DB*; therefore the squares on *AD*, *DB* are double of the squares on *AC*, *CD*.

Therefore etc. Q. E. D.