Inscribing Circle in Triangle

Theorem

In the words of Euclid:

In a given triangle to inscribe a circle.

(The Elements: Book $\text{IV}$: Proposition $4$)

Let $\triangle ABC$ be the given triangle.

Let $\angle ABC$ and $\angle ACB$ be bisected by $BD$ and $CD$ and let these lines join at $D$.

From $D$ construct the perpendiculars $DE, DF, DG$ to $AB, BC, AC$ respectively.

Draw the circle with radius $DE$ and center $D$.

This is the required circle.

We have that $\angle ABD = \angle CBD$ and $\angle BED = \angle BFD$, a right angle, and $BD$ is common.

$\triangle EBD = \triangle FBD$

So $DE = DF$.

For the same reason $DG = DF$.

So $DE = DF = DG$.

So the circle drawn with radius $DE$ will pass through $E, F$ and $G$.

From Line at Right Angles to Diameter of Circle it follows that $AB, AC, BC$ are tangent to the circle $EFG$.

Hence the result.

$\blacksquare$

This proof is Proposition $4$ of Book $\text{IV}$ of Euclid's The Elements.

1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{IV}$. Propositions
1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The incentre
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): incentre
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): incentre