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题目如图,在锐角三角形ABC中,∠BAC≠60°,过点B、C分别作三角形ABC的外接圆的切线BD、CE,且满足BD=CE=BC.直线DE与AB、AC的延长线分别交于点F、G.设CF与BD交于点M,CE与BG交于点N.证明:AM=AN.这是2014年全国高中数学联合竞赛加试的第二题.组委会提供的两种解法中,一种是多次利用角平分线性质定理得证,技巧性较强;另一种是利用余弦定理,通过计算线段长度求得.本文给出两种比较基础的证明方法,供读者参考.
The subject is shown in the figure, in the acute triangle ABC, ∠ BAC ≠ 60 °, and points B, C are the circumscribed circles of the circumscribed circle of triangle ABC BD, CE, and satisfy BD = CE = BC. The lines are respectively intersected with points F and G. Let CF and BD meet at point M and CE and BG meet at point N. Proof: AM = AN. This is the second question to be added in the 2014 national high school math joint competition test. One of the two solutions that will be provided is that it is proved by many times to use the theorem of bisector of angle and the skill is strong. The other one is calculated by using the cosine theorem and the length of the line segment.Two kinds of comparatively basic Proof methods for readers reference.