三角形ABC中,角A为直角,D为AC的中点,AE垂直DB于E,且BE=2DE,求证：BC=3AE

可设a=DE,则BE=2a△AED∽△BEA所以AE^2=BE*DE=2a^2AB^2 = AE^2+BE^2 = 2a^2+4a^2=6a^2AD^2 = AE^2 + DE^2 = 2a^2+a^2=3a^2AC=2ADAC^2=4AD^2=12a^2BC^2=AB^2+AC^2=6a^2+12a^2=18a^2BC^2/AE^2=18a^2/(2a^2)=9所以BC/AE...