如图,平面直角坐标系中,矩形ABCO的边OA在y正半轴上,OC在x正半轴上,点D是线段OC上一点,过点D作DE⊥AD交直线BC于点E,以A、D、E为顶点作矩形ADEF．（1）求证：△AOD∽△DCE；（2）若点A坐标为（0,4）,点C坐标为（7,0）．①当点D的坐标为（5,0）时,抛物线y=ax2+bx+c过A、F、B三点,求点F的坐标及a、b、c的值；②若点D（k,0）是线段OC上任意一点,点F
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(1) ∠AOD = ∠DCE = ∠ADE = 90°
∠CDE + ∠CED = 90°,∠OAD + ∠ODA = 90°
∠ODA + ∠CDE = 180° - ∠ADE = 180° - 90° = 90°
即∠CDE和∠ODA互余,∠CDE和∠CED也互余,∠ODA = ∠CED
△AOD∽△DCE
(2) B(7,4)
①OD = 5,CD = 2,OA = 4
△AOD∽△DCE
,CE/OD = CD/OA,CE/2 = 5/4CE = 5/2
E(7,5/2)
从F向AB做垂线,垂足G,显然△AFG与△DCE全等.F的横坐标=DC=2,F的纵坐标=A的纵坐标+GF=4+CE= 4+ 5/2 = 13/2
F(2,13/2)
过A:c = 4
过B:49a + 7b + c = 4
过F:4a + 2b + c = 13/2
解为：a = -1/4,b = 7/4,c = 4
②与①类似,DC= 7-k,OD = k
△AOD∽△DCE
,CE/OD = CD/OA,CE/k = (7 - k)/4CE = k(7-k)/4
从F向AB做垂线,垂足G,显然△AFG与△DCE全等.F的横坐标=DC=7-k,F的纵坐标=A的纵坐标+GF=4+CE= 4+ k(7-k)/4
F(7-k,4+ k(7-k)/4)
代入抛物线方程：4+ k(7-k)/4 = -(7-k)²/4 + 7(7 - k)/4 + 4
k(7-k)/4 = -(7-k)²/4 + 7(7 - k)/4
k(7-k) = -(7-k)² + 7(7 - k)
k(7-k) = (7 - k)(-7+k +7)
k(7-k) = k(7-k)
恒成立（注：k = 7时,C与D重合,无意义)
(3)存在,y = -x²/4 + nx/m + m