# Three Incircles in a Triangle:

What Is This About?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny

This is an extension of another result concerning three incircles in a triangle.

### Theorem

Let the incircle of ΔABC touch the side BC at point T. Let X be another point on BC. Consider incircles of triangles ABX and ACX. Then the internal common tangent of the two circles other than AX, passes through T. Additionally, both X and T lie on a circle constructed on the segment joining the incenters of triangle ABX and ACX as diameter.

(For a partial history of the problem, a context and the credits, refer to D. Grinberg's post (Theorems 1 and 2) at the CTK Exchange.)

### Proof

Below, I shall refer to the following configuration, where X lies to the right of T.

Let I_{B} and I_{C} be the incenters of triangles ABX and ACX, respectively; T_{B} and T_{C} are the points where their incircles touch AX. AX is thus an internal common tangent of the two circles. Assume their other internal common tangent S_{B}S_{C} crosses BC in point S. We want to show that

Obviously,

(1)

T_{B}T_{C} = S_{B}S_{C}.

If B' and C' are the points where the two incircles touch BC, then _{B}_{C},

(2)

S_{B}S_{C} = SS_{C} - SS_{B} = SC' - SB'.

On the other hand,

(3)

S_{B}S_{C} = T_{B}T_{C} = XT_{B} - XT_{C} = XB' - XC',

wherefrom

(4)

SB' + XB' = SC' + XC'.

And we conclude that the midpoints of segments SX and B'C' coincide, and therefore SB' = XC'. Now, as was shown by V. Zajic,

(5)

TX = T_{B}T_{C}.

From (3) and (5) we also have

(6)

TX + XC' = XB' = TB' + TX,

which implies XC' = TB', so that, finally, TB' = SB' and T = S.

Let's now consider the circle, say g, with diameter I_{B}I_{C}. I_{B} lies on the bisector of angle BXA, while I_{C} lies on the bisector of the supplementary angle CXA. Since the two bisectors are orthogonal, X lies on g. Since TS_{B}S_{C} is also tangent to the two circles, I_{B} and I_{C} lie on the bisectors of BTS_{B}S_{C} and CTS_{B}S_{C} - another pair of supplementary angles. The two are again orthogonal, and therefore T, too, lies on g.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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