# Proof: By Euclid

(related to Proposition: 4.04: Inscribing Circle in Triangle)

• Let the angles $ABC$ and $ACB$ have been cut in half by the straight lines $BD$ and $CD$ (respectively) [Prop. 1.9], and let them meet one another at point $D$, and let $DE$, $DF$, and $DG$ have been drawn from point $D$, perpendicular to the straight lines $AB$, $BC$, and $CA$ (respectively) [Prop. 1.12].
• And since angle $ABD$ is equal to $CBD$, and the right angle $BED$ is also equal to the right angle $BFD$, $EBD$ and $FBD$ are thus two triangles having two angles equal to two angles, and one side equal to one side - the (one) subtending one of the equal angles (which is) common to the (triangles) - (namely), $BD$.
• Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
• Thus, $DE$ (is) equal to $DF$.
• So, for the same (reasons), $DG$ is also equal to $DF$.
• Thus, the three straight lines $DE$, $DF$, and $DG$ are equal to one another.
• Thus, the circle drawn with center $D$, and radius one of $E$, $F$, or $G$,1 will also go through the remaining points, and will touch the straight lines $AB$, $BC$, and $CA$, on account of the angles at $E$, $F$, and $G$ being right angles.
• For if it cuts (one of) them then it will be a (straight line) drawn at right angles to a diameter of the circle, from its extremity, falling inside the circle.
• The very thing was shown (to be) absurd [Prop. 3.16].
• Thus, the circle drawn with center $D$, and radius one of $E$, $F$, or $G$, does not cut the straight lines $AB$, $BC$, and $CA$.
• Thus, it will touch them and will be the circle inscribed in triangle $ABC$.
• Let it have been (so) inscribed, like $FGE$ (in the figure).
• Thus, the circle $EFG$ has been inscribed in the given triangle $ABC$.
• (Which is) the very thing it was required to do.

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### References

1. Here, and in the following propositions, it is understood that the radius is actually one of $DE$, $DF$, or $DG$ (translator's note).