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题目与分析 1994年杭州市第六届“求是杯”初中数学竞赛的压轴题是这样的一道题:如图1,△ASC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC.AM与BN相交于点P.求证:∠BPM=45°. 由于条件给出的是线段的等量关系,求证的却是角度等式,二者不易沟通.但由于条件中有直角和线段相等,因此,既可想到构造等腰直角三角形,又可利用勾股定理去引伸出新的等量关系,于是有下面的两种途
Topics and Analysis The finale of the 6th “Qiusi Cup” junior high school mathematics competition in Hangzhou in 1994 was a question like this: Figure 1, ASC, ∠C=90°, point M on BC, and BM= AC, point N is on AC, and AN=MC.AM intersects with BN at point P. Proof: ∠BPM=45°. Because the condition gives the isometric relationship of line segments, the angle equation is verified, It is not easy to communicate. However, because there are right angles and line segments in the conditions, it is possible to think of constructing isosceles triangles and using the Pythagorean theorem to introduce new equal relations, so there are the following two ways