## An inscribed circle theorem

Posted September 1st, 2007 by Isoscel

In a triangle ABC if the inscribed circle is tangent in A_{1}, B_{1} and C_{1} to the sides BC, CA and AB then AA_{1}, BB_{1} and CC_{1} intersect.

**Proof:**

We use the fact that a point that is exterior to a circle has the tangents to it equal in length.

AB_{1} and AC_{1} are tangets to the circle with B_{1} and C_{1} on the circle, so:

AB_{1}=AC_{1}=a_{T}

In the same way:

BC_{1}=BA_{1}=b_{T}

CA_{1}=CB_{1}=c_{T}

To prove that AA_{1}, BB_{1} and CC_{1} intersect we will use Ceva's theorem.

We have to prove that:

This is equivalent with this simple identity: