# Homothety between In- and Excircles

A circle is inscribed in ΔABC. MN is the diameter perpendicular to the side AC. Let L be the intersection of BM with AC. Prove that

What if applet does not run? |

### References

- A. Soifer,
*Mathematics as Problem Solving*, Springer (April 28, 2009)

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

Copyright © 1996-2018 Alexander Bogomolny

### Solution

A circle is inscribed in ΔABC. MN is the diameter perpendicular to the side AC. Let L be the intersection of BM with AC. Prove that

What if applet does not run? |

Let EF (E on AB, F on BC) be the tangent to the incircle at M. Then EF||AC, making triangles ABC and EBF similar and, hence, homothetic from point B. Under that homothety, M is mapped to L. The incircle (I) of ΔABC is an excircle of ΔEBF. It is therefore mapped on the excircle (I_{b}) of ΔABC.

Just to summarize, L and N are the points of tangency of the incircle (I) and the excircle (I_{b}) of ΔABC with side AC.

We are going to use two facts:

- Two tangents from a point two a circle are equal in length.
- Two tangents - both either internal or external - to two circles are equal in length.

Note that the first implies the second.

Circles (I) and (I_{b}) touch AB at E and Y and BC at Z and W. By the second property,

We also have

AY = AL,

AX = AN,

CW = CL,

CN = CZ.

Combining all five gives

CL + CN = AL + AN.

Observe now that, depending on the relative positions of points L and N within the segment AC, either

(CN ± LN) + CN = AL + (AL ± LN).

so that 2CN = 2AL and, hence, CN = AL.

We actually may be more specific. Let's, as usual, denote the sidelength of ΔABC a, b, and c, and its semiperimeter

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

Copyright © 1996-2018 Alexander Bogomolny

71948767