---
book: 4
number: 5
id: "IV.5"
kind: "construction"
uses: ["[[book-1/proposition-10]]", "[[book-1/proposition-4]]", "[[book-3/proposition-31]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.5"
license: "CC-BY-SA-4.0"
---

# IV.5

*About a given triangle to circumscribe a circle*.

## Proof

Let *ABC* be the given triangle; thus it is required to circumscribe a circle about the given triangle *ABC*.

Let the straight lines *AB*, *AC* be bisected at the points *D*, *E* [[book-1/proposition-10|I. 10]], and from the points *D*, *E* let *DF*, *EF* be drawn at right angles to *AB*, *AC*; they will then meet within the triangle *ABC*, or on the straight line *BC*, or outside *BC*.

First let them meet within at *F*, and let *FB*, *FC*, *FA* be joined.

Then, since *AD* is equal to *DB*, and *DF* is common and at right angles, therefore the base *AF* is equal to the base *FB*. [[book-1/proposition-4|I. 4]]

Similarly we can prove that *CF* is also equal to *AF*; so that *FB* is also equal to *FC*; therefore the three straight lines *FA*, *FB*, *FC* are equal to one another.

Therefore the circle described with centre *F* and distance one of the straight lines *FA*, *FB*, *FC* will pass also through the remaining points, and the circle will have been circumscribed about the triangle *ABC*.

Let it be circumscribed, as *ABC*.

Next, let *DF*, *EF* meet on the straight line *BC* at *F*, as is the case in the second figure; and let *AF* be joined.

Then, similarly, we shall prove that the point *F* is the centre of the circle circumscribed about the triangle *ABC*.

Again, let *DF*, *EF* meet outside the triangle *ABC* at *F*, as is the case in the third figure, and let *AF*, *BF*, *CF* be joined.

Then again, since *AD* is equal to *DB*, and *DF* is common and at right angles, therefore the base *AF* is equal to the base *BF*. [[book-1/proposition-4|I. 4]]

Similarly we can prove that *CF* is also equal to *AF*; so that *BF* is also equal to *FC*; therefore the circle described with centre *F* and distance one of the straight lines *FA*, *FB*, *FC* will pass also through the remaining points, and will have been circumscribed about the triangle *ABC*.

Therefore about the given triangle a circle has been circumscribed. Q. E. F.

And it is manifest that, when the centre of the circle falls within the triangle, the angle *BAC*, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line *BC*, the angle *BAC*, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle *BAC*, being in a segment less than the semicircle, is greater than a right angle. [[book-3/proposition-31|III. 31]]