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.

Proof:

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:
AB1=AC1=aT
In the same way:
BC1=BA1=bT
CA1=CB1=cT

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

We have to prove that:

$$\frac{AC_1}{C_1B}\,\cdot\,\frac{BA_1}{A_1C}\,\cdot\,\frac{CB_1}{B_1A}=1$$

This is equivalent with this simple identity:

$$\frac{a_T}{b_T}\,\cdot\,\frac{b_T}{c_T}\,\cdot\,\frac{c_T}{a_T}=1$$

$<br />
\begin{picture}(500,380)<br />
\put(280,196){\circle(132)}</p>
<p>\put(80,90){A}<br />
\put(300,340){B}<br />
\put(370,130){C}</p>
<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 />
\put(335,115){c_T}</p>
<p>\put(350,210){A_1}<br />
\put(280,100){B_1}<br />
\put(200,240){C_1}</p>
<p>\put(100,100){\line(247,110)}<br />
\put(320,340){\line(-28,-211)}<br />
\put(360,140){\line(-130,102)}</p>
<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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