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苏科版《数学》七年级下册第29页例2:如图1,△ABC的角平分线BD,CE相交于点P,∠A=70°,求∠BPC的度数.今天我们就从这道题着手,探究例题的解题结论在问题变式后的有效应用,并通过例题的进一步拓展,达到解一题而通一类,举一反三的高效解题目的.一、例题的解题结论【解析】利用三角形内角和性质得∠ABC+∠ACB的度数,根据角平分线性质得∠1+∠2的度数,然后根据三角形内角
Example 2: As shown in Fig. 1, the angle bisector BD of ABC, CE intersects at point P, ∠A = 70 °, and the degree of BPC is obtained. From today This question starts with an exploration of the effective application of the solution to the problem of the problem after the variation of the problem and through the further expansion of the example to achieve the purpose of solving one problem through one and giving top priority to solving the problem. 【Resolution】 Using the angle and the nature of the triangle ∠ ABC + ∠ ACB degrees, according to the nature of the angle bisector obtained ∠ 1 + ∠ 2 degree, and then according to the triangle angle