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let triangle with AB - BC =AC / \sqrt2

Source: 2009 Sharygin Geometry Olympiad Final Round problem 6 grade 9

July 26, 2018

geometryangles

Problem Statement

Given triangle

ABC

such that

AB- BC = \frac{AC}{\sqrt2}

. Let

M

be the midpoint of

AC

, and

N

be the foot of the angle bisector from

B

. Prove that

\angle BMC + \angle BNC = 90^o

.
(A.Akopjan)