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原题如图1,在正方形ABCD中,E、F分别是边BC、CD上的点,∠EAF=45°.求证:EF=BE+FD.本题是一道经典几何题,其知识丰富,思想方法独特,具有较强的拓展价值,对它进行一题多证、一题多得、一题多变、一题多用的处理.一题多证证法一延长EB至G,使BG=DF,连结AG(如图2).因为四边形ABCD是正方形,所以BA=DA,∠D=∠DAB=∠ABE=∠ABG=90°.所以△ABG≌△ADF,所以AG=AF,∠1=∠2.因为∠EAF=45°,所以∠1+∠3=∠2+∠3=45°.又AE是公共边,所以△AEG≌△AEF,所以EG=EF,即EF=BE+FD.
The original title shown in Figure 1, in the square ABCD, E, F, respectively, the edge BC, points on the CD, ∠ EAF = 45 °. Confirmation: EF = BE + FD. The problem is a classical geometry, its knowledge-rich, The method is unique, has a strong expansion value, to carry out a question more than one card, a question more, a question changeable, a multi-purpose treatment. , Link AG (Figure 2). Because the quadrilateral ABCD is a square, so BA = DA, ∠D = ∠DAB = ∠ABE = ∠ABG = 90 °. So △ ABG≌ △ ADF, so AG = AF, ∠1 = ∠2. Because ∠EAF = 45 °, so ∠1 + ∠3 = ∠2 + ∠3 = 45 °. And AE is the public side, so △ AEG≌ △ AEF, so EG = EF, ie EF = BE + FD .