---
book: 6
number: 3
id: "VI.3"
kind: "theorem"
uses: ["[[book-1/proposition-29]]", "[[book-1/proposition-6]]", "[[book-6/proposition-2]]", "[[book-5/proposition-11]]", "[[book-5/proposition-9]]", "[[book-1/proposition-5]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.3"
license: "CC-BY-SA-4.0"
---

# VI.3

*If an angle of a triangle be bisected and the straight line cutting the angle cut the base also*, *the segments of the base will have the same ratio as the remaining sides of the triangle; and*, *if the segments of the base have the same ratio as the remaining sides of the triangle*, *the straight line joined from the vertex to the point of section will bisect the angle of the triangle*.

## Proof

Let *ABC* be a triangle, and let the angle *BAC* be bisected by the straight line *AD*; I say that, as *BD* is to *CD*, so is *BA* to *AC*.

For let *CE* be drawn through *C* parallel to *DA*, and let *BA* be carried through and meet it at *E*.

Then, since the straight line *AC* falls upon the parallels *AD*, *EC*, the angle *ACE* is equal to the angle *CAD*. [[book-1/proposition-29|I. 29]]

But the angle *CAD* is by hypothesis equal to the angle *BAD*; therefore the angle *BAD* is also equal to the angle *ACE*.

Again, since the straight line *BAE* falls upon the parallels *AD*, *EC*, the exterior angle *BAD* is equal to the interior angle *AEC*. [[book-1/proposition-29|I. 29]]

But the angle *ACE* was also proved equal to the angle *BAD*; therefore the angle *ACE* is also equal to the angle *AEC*, so that the side *AE* is also equal to the side *AC*. [[book-1/proposition-6|I. 6]]

And, since *AD* has been drawn parallel to *EC*, one of the sides of the triangle *BCE*, therefore, proportionally, as *BD* is to *DC*, so is *BA* to *AE*.

But *AE* is equal to *AC*; [[book-6/proposition-2|VI. 2]] therefore, as *BD* is to *DC*, so is *BA* to *AC*.

Again, let *BA* be to *AC* as *BD* to *DC*, and let *AD* be joined; I say that the angle *BAC* has been bisected by the straight line *A.D*.

For, with the same construction, since, as *BD* is to *DC*, so is *BA* to *AC*, and also, as *BD* is to *DC*, so is *BA* to *AE* : for *AD* has been drawn parallel to *EC*, one of the sides of the triangle *BCE*: [[book-6/proposition-2|VI. 2]] therefore also, as *BA* is to *AC*, so is *BA* to *AE*. [[book-5/proposition-11|V. 11]]

Therefore *AC* is equal to *AE*, [[book-5/proposition-9|V. 9]] so that the angle *AEC* is also equal to the angle *ACE*. [[book-1/proposition-5|I. 5]]

But the angle *AEC* is equal to the exterior angle *BAD*, [[book-1/proposition-29|I. 29]] and the angle *ACE* is equal to the alternate angle *CAD*; [*id*.] therefore the angle *BAD* is also equal to the angle *CAD*.

Therefore the angle *BAC* has been bisected by the straight line *AD*.

Therefore etc. Q. E. D.