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证明含三角函数的几何等式,不少同学感到难以下手,如应用锐角三角函数的定义,将式子中的三角函数转换为两线段的比,从而将问题转化为线段的等比(积),常可迎刃而解。 例1 如图1,△ABC中,以BC为直径的半圆分别和AB、AC交于D、E.求证:DE=BCcosA (1994,西安市中考题) 分析:连BE,则∠BEC=90°,△ABE为直角三角形,从而命题转化为证明DE=BC·AE/AB,即证DE/BC=AE/AB. 为此,可证△ADE∽△ACB. 由∠ADE=∠ACB,∠A=∠A.命题获证.
Prove that geometric equations containing trigonometric functions, many students feel difficult to start, such as the application of the definition of acute angle trigonometric function, the trigonometric function in the equation is converted to a ratio of two lines, so as to convert the problem into a line segment ratio (product) Can often be solved. Example 1 As shown in Figure 1, in △ABC, semicircles with BC diameter and AB, AC are submitted to D, E. Proof: DE=BCcosA (1994, Xi’an exam) Analysis: Even BE, then BEC=90 °, △ ABE is a right-angle triangle, so that the proposition is converted to prove that DE = BC · AE / AB, that is, proof DE / BC = AE / AB. To this end, can be proved △ ADE △ △ ACB. By ∠ ADE = ∠ ACB, ∠ A=∠A. Proposal certification.