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
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References
- A. Soifer, Mathematics as Problem Solving, Springer (April 28, 2009)
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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 (Ib) of ΔABC.
Just to summarize, L and N are the points of tangency of the incircle (I) and the excircle (Ib) 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 (Ib) touch AB at E and Y and BC at Z and W. By the second property,
< class="shift"> XY = ZW. >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
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