Date: Sat, 25 Jan 1997 14:05:51 -0500

From: Alexander Bogomolny

Ronald:

Let the triangle be ABC and denote as A_{1}, B_{1}, and C_{1}
bases of the altitudes AA_{1}, BB_{1}, and CC_{1}, respectively.
Considering six triangles ABA_{1}, ABB_{1}, ACA_{1}, ACC_{1}, BCC_{1}, and BCB_{1},
you may conclude that:

_{1}/CC

_{1}=AB

_{1}/BB

_{1}

cotB=BC

_{1}/CC

_{1}=BA

_{1}/AA

_{1}

cotC=CB

_{1}/BB

_{1}=CA

_{1}/AA

_{1}

Note that, for example, CC_{1}=2S/c, where S is the area of the triangle. Therefore,
c/CC_{1}=c^{2}/2S. Similar identities hold for a and b. Therefore,

^{2}/2S

cotB+cotC=a

^{2}/2S

cotC+cotA=b

^{2}/2S

Taking pairwise differences we obtain

^{2}-a

^{2})/2S

cotB-cotC=(c

^{2}-b

^{2})/2S

cotC-cotA=(a

^{2}-c

^{2})/2S

Rewrite your identity as

^{2}(cotC-cotB)+b

^{2}(cotA-cotC)+c

^{2}(cotB-cotA)=0

which becomes

^{2}(b

^{2}-c

^{2})+b

^{2}(c

^{2}-a

^{2})+c

^{2}(a

^{2}-b

^{2})]/2S=0

which is of course true.

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