---
book: 4
number: 2
id: "IV.2"
kind: "construction"
uses: ["[[book-3/proposition-16]]", "[[book-1/proposition-23]]", "[[book-3/proposition-32]]", "[[book-1/proposition-32]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.2"
license: "CC-BY-SA-4.0"
---

# IV.2

*In a given circle to inscribe a triangle equiangular with a given triangle*.

## Proof

Let *ABC* be the given circle, and *DEF* the given triangle; thus it is required to inscribe in the circle *ABC* a triangle equiangular with the triangle *DEF*.

Let *GH* be drawn touching the circle *ABC* at *A* [[book-3/proposition-16|III. 16, Por.]]; on the straight line *AH*, and at the point *A* on it, let the angle *HAC* be constructed equal to the angle *DEF*, and on the straight line *AG*, and at the point *A* on it, let the angle *GAB* be constructed equal to the angle *DFE*; [[book-1/proposition-23|I. 23]] let *BC* be joined.

Then, since a straight line *AH* touches the circle *ABC*, and from the point of contact at *A* the straight line *AC* is drawn across in the circle, therefore the angle *HAC* is equal to the angle *ABC* in the alternate segment of the circle. [[book-3/proposition-32|III. 32]]

But the angle *HAC* is equal to the angle *DEF*; therefore the angle *ABC* is also equal to the angle *DEF*.

For the same reason the angle *ACB* is also equal to the angle *DFE*; therefore the remaining angle *BAC* is also equal to the remaining angle *EDF*. [[book-1/proposition-32|I. 32]]

Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle. Q. E. F.