In triangle ABC, with <BAC=2<ACB, let AD be inner bisector of <BAC. Let denote M, N the midpoints of AC and BD. It turned out that <BAC=2<MNC; find the angles of the triangle.
Proof
Obviously DM_|_AC and <CAD=<CNM, thus ANDM is cyclic, AN_|_BC, so it is angle bisector of <BAD. So, if <ACD=<CAD=2x, <DAN=x and <ACD+<CAD+<DAN=90=5x, x=18, thus <A=36, <B=<C=72.
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