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/etc/nginx/nginx.conf.求解一道初二数学题,急!要过程!现在马上立刻!啊&啊啊&
血刃伤心b26
(1)【结论:四边形BECF为菱形】∵EF⊥BC又 ∠ACB=90°,即AC⊥BC∴ AC//EF (垂直于同一直线的两条直线平行)又,BD=CD则,DE为Rt△ABC的中位线,即有,BE=AE=CE (直角三角形斜边中点与直角顶点的连线等于斜边的一半)又CF=AE而 由垂直平分线性质有,BF=CF故可得,BF=CF=CE=BE,∴四边形BECF为菱形当四边形BECF为正方形时,∠FBA = 90°而,BC平分∠FBA (正方形的性质)则 ∠CBA = (1/2)*90° = 45°在Rt△ABC中,∠A = 90° - 45° = 45°∴当四边形BECF为正方形时,∠A=45°
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1、菱形 2、∠A=45°
看不见么???是图片。。。
(1)【结论:四边形BECF为菱形】
又 ∠ACB=90°, 即AC⊥BC
(垂直于同一直线的两条直线平行)
则,DE为Rt△ABC的中位线,
即有,BE=AE=CE ...
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