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:
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,
cotB+cotC=a2/2S
cotC+cotA=b2/2S
Taking pairwise differences we obtain
cotB-cotC=(c2-b2)/2S
cotC-cotA=(a2-c2)/2S
Rewrite your identity as
which becomes
which is of course true.
| 40620616 |
