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在平面几何中,有如下一个著名的问题(汤普森问题[1]):在△ABC中,AB=AC,∠BAC=20°,D、E分别在AC和AB上,∠DBC=60°,∠ECB=50°.求∠BDE的度数.图1受文献[1]的启发,本文给出以下几种解法,供同学们鉴赏.解法1(用“三角形外心”解)如图1所示,以B为圆心,BC为半径作圆弧交AC于点F,连接BF,EF,则∠CBF=180°-2∠BCA=20°.于是∠ABF=∠ABC-∠CBF=60°∴△BEF是正三角形.即FB=FE.又∠FDB=40°=∠FBD.则FB=FD.
In plane geometry, there is a famous problem (Thompson problem [1]): In ABC, AB = AC, BAC = 20 °, D, E on AC and AB respectively, ∠DBC = 60 °, ∠ECB = 50 °. Find the degree of ∠BDE. Figure 1 Inspired by the literature [1], this paper gives the following solutions for students to appreciate. Solution 1 (with “triangular outer ” solution) 1, with B as the center of the circle, BC for the radius of the arc AC at point F, connecting BF, EF, then ∠ CBF = 180 ° -2 ∠ BCA = 20 ° So ∠ ABF = ∠ ABC-∠ CBF = 60 ° ∴ △ BEF is a regular triangle, that is, FB = FE. And ∠ FB = 40 ° = ∠ FBD. Then FB = FD.