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本文介绍一道几何题的四种解法,以开阔解题视野,提高创造新思维的能力,供解题参考.问题在梯形ABCD中,AD∥BC,∠B+∠C=90°,AD=7,BC=15,E、F分别是AD、BC的中点,求EF的长.分析注意到题设中有“90°”和“中点”的条件,设法构造出直角三角形,应用斜边上的中线的性质.方法一延长法因为∠B+∠C=90°,分别延长BA,CD交于点P,组成Rt△PAD和Rt△PBC,用直角三角形斜边上的中线性质解答.解分别延长BA,CD交于点P,如图1所示.∵∠B+∠C=90°,
This paper introduces a solution of four geometric problems in order to broaden the field of view to solve problems and improve the ability to create new thinking for the problem-solving reference problem in trapezoidal ABCD, AD∥BC, ∠B + ∠C = 90 °, AD = 7, BC = 15, E and F are the midpoints of AD and BC, respectively, and find the length of EF. The analysis notices that there are conditions of “90 ° ” and "midpoint Applying the nature of the center line on the hypotenuse. For the extension method, because BA + ∠C = 90 °, extend BA and CD respectively to point P and form Rt △ PAD and Rt △ PBC. Using the center line on the hypotenuse of right triangle Solution. Resolve the extension of BA, CD at point P, as shown in Figure 1. ∵∠ B + ∠ C = 90 °,