# Proving that the sides of a triangle are perpendicular to the incircle

I wonder how I could prove $AC\perp OD$ when the incircle of $\triangle ABC$ has been constructed as shown in the image below without using the argument that $AC$ is a tangent of the circle.

Using the following definition:

the incircle of a triangle is the circle which has exactly one common point with each side of the triangle.

I was trying something like this, but it seems like circular reasoning:

Since $|DO|=|OF|=r$ and $O$ is the interesection point of the 3 angular bisectors. And the angular bisector $AO$ is the set of point which are located at the same distance from $AC$ and $AB$ we can conclude that $|DO|$ is the distance of $O$ to $AC$ which results in perpendicularity.

(however, why does the definition above implies $O$ to be the intersection of the angular bisectors...)

• That seems like a poor definition for the incircle, since it assumes both that such a circle exists and also that there is only one such circle. – Rory Daulton Mar 23 '15 at 21:09

Consider just the line $AC$ in your diagram and the arc of the circle going through point $D$. $AC$ meets $OD$ at a distance $r$ from $O$; that is, $OD = r$.
Assume that $AC$ is not perpendicular to $OD$ and choose the smaller of these unequal angles $\angle ODA, \angle ODC$. Without loss of generality say this is $\angle ODA$. Now drop a perpendicular from $O$ $OX$ to line $AC$, which under these assumptions will lie between $D$ and C$. Next, consider the point$D'$formed by going a distance$DX$from$X$along line$XC$toward$C$. Consider the triangles$OXD$and$OXD'$. These are congruent triangles by SAS since sll right angles are equal. Thus$OD = OD'$and$D'$lies on the circle. So if$\angle ODA \neq 90^\circ$then$AC$meets the circle in two points. But the definitiion of the incircle demands that it meets at only 1 point. so $$AC\perp OD$$ Geometry already has the theorem that a line tangent to a circle is perpendicular to a radius drawn to the intersection point. Or to quote a textbook, Theorem 11-1-1 in Geometry by Burger et al., If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency. Remember, that a tangent line to a circle is defined as a line that intersects the circle exactly once. So if the circle "has exactly one common point with each side of the triangle" then it is tangent to each side. That is all that is needed here. If you want to see a proof of that theorem, see this post on this site. • Ok, but I wonder if it would be possible to prove this special case without resorting to this known theorem. That's actually what the question was about (as denoted in the question) – dietervdf Mar 23 '15 at 21:15 • If you are asking for a proof of that theorem then you are duplicated a previous question on this site, as I just noted in my answer. Is some detail of those answers giving you some problems? – Rory Daulton Mar 23 '15 at 21:19 You know that the incircle intersects$AC$, by definition. If$OD$is not orthogonal to$AC$then it must intersect$AC$in another point$D'$, where$D'$is the symmetric of$D$with respect to the feet of the perpendicular from$O$to$AC$. Thus the circle intersects$AC$in two points. Contradiction. The contradiction came from the assumption that$OD$is not orthogonal to$AC$. This assumptioin is false, so$OD \perp AC\$.