解法1 设AB =" AC" = 1,CD = x,则0<x<1,BC =,AD = 1-x. 在Rt△ABD中,BD2 = AB2 + AD2 =" 1" +(1-x)2 = x2-2x + 2. 由已知可得 Rt△ABD∽Rt△ECD, ∴,即,从而, ∴,0<x<1, (1)若BD是AC的中线,则CD =" AD" =" x" =,得. (2)若BD是∠ABC的角平分线,则,得,解得, ∴. (3)若,则有 3x2-10x + 6 = 0,解得∈(0,1), ∴,表明随着点D从A向C移动时,BD逐渐增大,而CE逐渐减小,的值则随着D从A向C移动而逐渐增大. 解法2 设AB =" AC" = 1,∠ABD = a,则 BC =,∠CBE = 45°-a. 在Rt△ABD中,有; 在Rt△BCE中,有 CE =" BC·" sin∠CBE =sin(45°-a). 因此.下略…… 解法3 (1)∵∠A =∠E = 90°,∠ADB =∠CDE,∴△ADB∽△EDC,∴. 由于D是中点,且AB = AC,知AB =" 2" AD,于是 CE =" 2" DE. 在Rt△ADB中,BD =. 在Rt△CDE中,由 CE2 + DE2 = CD2,有 CE2 +CE2 = CD2,于是. 而 AD = CD,所以. (2)如图,延长CE、BA相交于点F.∵BE是∠ABC的平分线,且BE⊥CF,∴△CBE≌△FBE,得 CE = EF,于是CF =" 2" CE.又∠ABD +∠ADB =∠CDE +∠FCA = 90°,且∠ADB =∠CDE, ∴∠ABD =∠FCA,进而有△ABD≌△ACF,得 BD =" 2" CE,. (3)的值的取值范围为≥1.下略…… |