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题目Rt△ABC中,∠C=90°,以斜边AB为边向外作正方形ABDE,且正方形对角线交于点O,连OC,已知AC=5,OC=6槡2,求直角边BC的长.思路一:模型图联想法(即美国总统菲尔德勾股定理证法典型图,如图2)由Rt△ABC、Rt△ABD,还缺一个“竖立”右边的以BD为斜边的直角三角形BDF!于是补全之(延长CB,过点D作CB垂线,垂足为F),见图3.图2图3Rt△ABC中,∠1+∠3=90°,又∠1+∠2=180°-90°=90°,得∠2=();
Problem Rt △ ABC, ∠ C = 90 °, with the beveled edge AB outward for the square ABDE, and the square diagonal intersect at point O, with OC, known AC = 5, OC = 6 槡 2, seeking The right-angle side of the BC long. Way of thinking: the model figure association idea (that is, the United States President Field Pythagorean theorem proof of law map, shown in Figure 2) by Rt △ ABC, Rt △ ABD, but also a lack of “erect” to the right BD is a hypotenuse of the right triangle BDF! Then complemented (extended CB, D over the point for the CB vertical, vertical foot F), shown in Figure 3. Figure 2 Figure 3Rt △ ABC, ∠1 + ∠3 = 90 °, and ∠1 + ∠2 = 180 ° -90 ° = 90 °, so ∠2 = ();