△ABC是等腰直角三角形,∠ACB=90°,过BC的中点D作DE⊥AB于E,连结CE设BE=DE=a,则BD=DC=(√2)aBC=AC=(2√2)a,BC^2=AC^2=8a^2AB=(√2)BC=(√2)*(2√2)a=4aAE=3a由余弦定理,得CE^2=BE^2+BC^2-2BE*BC*cosB=a^2+8a^2-2a*(2...