Subject: Re: Trigonometry in a triangle
Date: Sat, 25 Jan 1997 14:05:51 -0500
From: Alexander Bogomolny

Ronald:

Let the triangle be ABC and denote as A1, B1, and C1 bases of the altitudes AA1, BB1, and CC1, respectively. Considering six triangles ABA1, ABB1, ACA1, ACC1, BCC1, and BCB1, you may conclude that:

cotA=AC1/CC1=AB1/BB1
cotB=BC1/CC1=BA1/AA1
cotC=CB1/BB1=CA1/AA1

Note that, for example, CC1=2S/c, where S is the area of the triangle. Therefore, c/CC1=c2/2S. Similar identities hold for a and b. Therefore,

cotA+cotB=c2/2S
cotB+cotC=a2/2S
cotC+cotA=b2/2S

Taking pairwise differences we obtain

cotA-cotB=(b2-a2)/2S
cotB-cotC=(c2-b2)/2S
cotC-cotA=(a2-c2)/2S

Rewrite your identity as

a2(cotC-cotB)+b2(cotA-cotC)+c2(cotB-cotA)=0

which becomes

[a2(b2-c2)+b2(c2-a2)+c2(a2-b2)]/2S=0

which is of course true.

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