perplexus.info :: Geometry : Bisector Inequality

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|.

See The Solution Submitted by Bractals

Rating: 4.0000 (1 votes)

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Solution

Comment 1 of 1

Let O be the point of intersection of AD and CE and therefore the centre of the in-circle with radius r, say. Let P and Q be the feet of the perpendiculars from O to the sides BC and AB respectively.

   |AC| = |AQ| + |CP|                (using tangent lengths to the in-circle)
            = |AE| + r cot(BEC) + |CD| + r cot(BDA)
            = |AE| + |CD| + r[cot(BEC) + cot(BDA)]                          (1)

So it only remains to show that cot(BEC) + cot(BDA) > 0 for B > pi/3.

cot(BEC) + cot(BDA)    = cot(pi - B - C/2) + cot(pi - B - A/2)

                                    = - cot(B + C/2) - cot(B + A/2)

                        = - cos(B + C/2) sin(B + A/2) + cos(B + A/2) sin(B + C/2)
                                                sin(B + A/2) sin(B + C/2)

                        =         _____- sin(2B + A/2 + C/2)__________
                                    0.5[cos(A/2 - C/2) - cos(2B + A/2 + C/2)]

                        =          ____- 2 sin(3B/2 + pi/2)______   (since A + B + C = pi)
                                    cos(A/2 - C/2) - cos(3B/2 +pi/2)

                        =          ____- 2 cos(3B/2)______
                                    cos(A/2 - C/2) + sin(3B/2)

If B > pi/3, then 3B/2 > pi/2 and cos(3B/2) < 0, so the minimum value of the above expression will occur when A = C, giving:

            cot(BEC) + cot(BDA)      >          - 2 cos(3B/2)
                                                            1 + sin(3B/2)

                                                >          0

Equation (1) then gives:    |AE| + |CD| < |AC|

Posted by Harry on 2010-08-23 23:00:34

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