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 AN = CL.

Solution

References

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

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Copyright © 1996-2012 Alexander Bogomolny

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:

1. Two tangents from a point two a circle are equal in length.
2. 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

Combining all five gives

Observe now that, depending on the relative positions of points L and N within the segment AC, either CN + LN = CL and AL + LN = AN, or CN - LN = CL and AL - LN = AN. Make a substitution:

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 p = (a + b + c)/2. Then, as we in fact know, CN = p - c = AL, AN = p - a = CL such that LN = |a - c|.

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Copyright © 1996-2012 Alexander Bogomolny