问题人教版数学八年级下册教科书第133页的第15题:如图1,四边形ABCD是正方形,点E是边BC的中点,∠AEF=90°,EF交正方形外角的平分线CF于F.求证:AE=EF.图1图2证明在AB上任取一点M,使AM=EC,连结ME,如图2.因为∠MAE=∠FEC,∠AME=∠ECF,所以△AME≌△ECF故AE=EF.这是按教材提示做辅 Question PEP mathematics grade eight books textbooks on page 133 of the 15th title: Figure 1, quadrilateral ABCD is a square, point E is the middle point of the side BC, ∠ AEF = 90 °, EF square corner of the intersection of the external bifurcation CF Proof in F.: AE = EF. Figure 1 Figure 2 proves that any point on AB to take a M, so that AM = EC, link ME, as shown in Figure 2. Because ∠ MAE = ∠ FEC, ∠ AME = ∠ ECF, so △ AME ≌ △ECF so AE=EF. This is supplemented by teaching materials