|To inscribe a circle in a given triangle.|
|Let ABC be the given triangle.
It is required to inscribe a circle in the triangle ABC.
|Bisect the angles ABC and ACB by the straight lines BD and CD, and let these meet one another at the point D. Draw DE, DF, and DG from D perpendicular to the straight lines AB, BC, and CA.||I.9|
|Now, since the angle ABD equals the angle CBD, and the right angle BED also equals the right angle BFD, EBD and FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that opposite one of the equal angles, which is BD common to the triangles, therefore they will also have the remaining sides equal to the remaining sides, therefore DE equals DF.||I.26|
|For the same reason DG also equals DF.
Therefore the three straight lines DE, DF, and DG equal one another. Therefore the circle described with center D and radius one of the straight lines DE, DF, or DG also passes through the remaining points and touches the straight lines AB, BC, and CA, because the angles at the points E, F, and G are right.
|For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its end will be found to fall within the circle, which was proved absurd, therefore the circle described with center D and radius one of the straight lines DE, DF, or DG does not cut the straight lines AB, BC, and CA Therefore it touches them, and is the circle inscribed in the triangle ABC.||III.16|
|Let it be inscribed as FGE.
Therefore the circle EFG has been inscribed in the given triangle ABC.
|The incircle is a circle tangent to the three lines AB, BC, and AC. If these three lines are extended, then there are three other circles also tangent to them, but outside the triangle. They are called the excircles.
The points on the internal angle bisector AD are equidistant from the two sides of the triangle AB and AC. The line KL is perpendicular to AD at A, and the points on it are also equidistant from the extended sides AB and AC. The line B'C' is called the external angle bisector at A. Whereas the incenter D lies at the confluence of the three internal angle bisectors, the excenter B' lies at the confluence of two external angle bisectors AB' and CB' and one internal angle bisector BB'. Likewise for the other two excenters A' and C'.
Let D be the incenter of the triangle ABC, and let DE, DF, and DG be perpendicular lines drawn to the sides as in Euclid's proof. These three lines are radii of the incircle, and therefore have length r, the inradius. The triangle ABD has base AB and height r, so its area is r AB/2. Likewise, the area of triangle BCD is r BC/2, and the area of triangle CAD is r CA/2. Adding these together we find the area of triangle ABC is r (AB + BC + CA)/2. Therefore we have
an interesting result in itself.
|Now let A' be the excenter on the bisector of the internal angle at A. Let A'E', A'F', and A'G' be the perpendiculars drawn from A' to the sides of the triangle. They are radii of the excircle of length rA. Triangle ABA' has base AB and height A'E', so its area is rA AB/2. Likewise, the area of triangle BCA' is rA BC/2, and the area of triangle CAA' is rA CA/2. Triangle ABC is the sum of triangles ABA' and ACA' minus triangle BCA', so its area is rA (AB + AC CA)/2 which equals rA(s A).|
From the other excircles we get two more equations. We then have
There are other relationships among these radii, for instance,
but let's stop here to go on to the circumcircle of a triangle.
Next proposition: IV.5