An inscribed circle theorem

Posted September 1st, 2007 by Isoscel

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=a_T
In the same way:
BC₁=BA₁=b_T
CA₁=CB₁=c_T

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:

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

$\begin{picture}(500,380) \put(280,196){\circle(132)} \put(80,90){A} \put(300,340){B} \put(370,130){C} \put(180,95){a_T} \put(140,165){a_T} \put(335,275){b_T} \put(250,290){b_T} \put(360,170){c_T} \put(335,115){c_T} \put(350,210){A_1} \put(280,100){B_1} \put(200,240){C_1} \put(100,100){\line(247,110)} \put(320,340){\line(-28,-211)} \put(360,140){\line(-130,102)} \put(100,100){\line(220,240)} \put(100,100){\line(260,40)} \put(360,140){\line(-40,200)} \end{picture}$

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An inscribed circle theorem

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