To circumscribe a triangle equiangular with a given triangle about a given circle.

Let *ABC* be the given circle, and *DEF* the given triangle.

It is required to circumscribe a triangle equiangular with the triangle *DEF* about the circle *ABC.*

Produce *EF* in both directions to the points *G* and *H.* Take the center *K* of the circle *ABC,* and draw a radius *KB* at random. On the straight line *KB* and at the point *K* on it, construct the angle *BKA* equal to the angle *DEG,* and the angle *BKC* equal to the angle *DFH.* Through the points *A,* *B,* and *C* draw *LAM, MBN,* and *NCL* touching the circle *ABC.*

Now, since *LM, MN,* and *NL* touch the circle *ABC* at the points *A, B,* and *C,* and *KA, KB,* and *KC* have been joined from the center *K* to the points *A, B,* and *C,* therefore the angles at the points *A, B,* and *C* are right.

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

But the sum of the angles *DEG* and *DEF* also equals two right angles, therefore the sum of the angles *AKB* and *AMB* equals the sum of the angles *DEG* and *DEF,* of which the angle *AKB* equals the angle *DEG,* therefore the remaining angle *AMB* equals the remaining angle *DEF.*

Similarly it can be proved that the angle *LNB* also equals the angle *DFE,* therefore the remaining angle *MLN* equals the angle *EDF.*

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

Therefore a triangle equiangular with the given triangle has been circumscribed about a given circle.

Q.E.F.