---
book: 4
number: 3
id: "IV.3"
kind: "construction"
uses: ["[[book-3/proposition-1]]", "[[book-1/proposition-23]]", "[[book-3/proposition-16]]", "[[book-3/proposition-18]]", "[[book-1/proposition-13]]", "[[book-1/proposition-32]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:4.prop.3"
license: "CC-BY-SA-4.0"
---

# IV.3

*About a given circle to circumscribe a triangle equiangular with a given triangle*.

## Proof

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

Let *EF* be produced in both directions to the points *G*, *H*, let the centre *K* of the circle *ABC* be taken [[book-3/proposition-1|III. 1]], and let the straight line *KB* be drawn across at random; on the straight line *KB*, and at the point *K* on it, let the angle *BKA* be constructed equal to the angle *DEG*, and the angle *BKC* equal to the angle *DFH*; [[book-1/proposition-23|I. 23]] and through the points *A*, *B*, *C* let *LAM*, *MBN*, *NCL* be drawn touching the circle *ABC*. [[book-3/proposition-16|III. 16, Por.]]

Now, since *LM*, *MN*, *NL* touch the circle *ABC* at the points *A*, *B*, *C*, and *KA*, *KB*, *KC* have been joined from the centre *K* to the points *A*, *B*, *C*, therefore the angles at the points *A*, *B*, *C* are right. [[book-3/proposition-18|III. 18]]

And, since the four angles of the quadrilateral *AMBK* are equal to four right angles, inasmuch as *AMBK* is in fact divisible into two triangles, and the angles *KAM*, *KBM* are right, therefore the remaining angles *AKB*, *AMB* are equal to two right angles.

But the angles *DEG*, *DEF* are also equal to two right angles; [[book-1/proposition-13|I. 13]] therefore the angles *AKB*, *AMB* are equal to the angles *DEG*, *DEF*, of which the angle *AKB* is equal to the angle *DEG*; therefore the angle *AMB* which remains is equal to the angle *DEF* which remains.

Similarly it can be proved that the angle *LNB* is also equal to the angle *DFE*; therefore the remaining angle *MLN* is equal to the angle *EDF*. [[book-1/proposition-32|I. 32]]

Therefore the triangle *LMN* is equiangular with the triangle *DEF*; and it has been circumscribed about the circle *ABC*.

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