Bisector Inequality (Posted on 2010-08-19)

In ΔABC the bisector of /BAC intersects side BC in point D 
and the bisector of /ACB intersects side AB in point E. 
Prove that

        /ABC > 60°  ==>  |AE| + |CD| < |AC|.

	Submitted by Bractals
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Solution:	(Hide)
Let a, b, and c denote the lengths of the sides of ΔABC opposite vertices A, B, and C respectively.         /ABC > 60° ==> cos(/ABC) < 1/2 ==> (a2 + c2 - b2)/(2ac) < 1/2 ==> a2 + c2 < ac + b2 ==> a2 + ab + bc + c2 < ab + ac + b2 + bc ==> a(a + b) + c(b + c) < (a + b)(b + c) ==> ab/(b + c) + bc(a + b) < b ==> |CD| + |AE| < |AC|.

	Subject	Author	Date
	Solution	Harry	2010-08-23 23:00:34