An inscribed circle theorem

In a triangle ABC if the inscribed circle is tangent in A1, B1 and C1 to the sides BC, CA and AB then AA1, BB1 and CC1 intersect.


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

AB1 and AC1 are tangets to the circle with B1 and C1 on the circle, so:
In the same way:

To prove that AA1, BB1 and CC1 intersect we will use Ceva's theorem.

We have to prove that:


This is equivalent with this simple identity:


$<br />
\begin{picture}(500,380)<br />
<p>\put(80,90){A}<br />
\put(300,340){B}<br />
<p>\put(180,95){a_T}<br />
\put(140,165){a_T}<br />
\put(335,275){b_T}<br />
\put(250,290){b_T}<br />
\put(360,170){c_T}<br />
<p>\put(350,210){A_1}<br />
\put(280,100){B_1}<br />
<p>\put(100,100){\line(247,110)}<br />
\put(320,340){\line(-28,-211)}<br />
<p>\put(100,100){\line(220,240)}<br />
\put(100,100){\line(260,40)}<br />
\put(360,140){\line(-40,200)}<br />
\end{picture}<br />

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