Answer (Continued) to Follow-up: a trigonometric identity at CTK Exchange

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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