题目如图,△ABC\triangle ABC△ABC内接于⊙O\odot O⊙O,∠BAC=75∘\angle BAC=75^{\circ}∠BAC=75∘,CDCDCD为⊙O\odot O⊙O的直径,且与△ABC\triangle ABC△ABC的边ABABAB交于点EEE,若AO⊥CDAO\bot CDAO⊥CD,BE=2BE=2BE=2,则BCBCBC的长是（ ）A.434\sqrt{3}43B.626\sqrt{2}62C.32+63\sqrt{2}+\sqrt{6}32+6D.33+223\sqrt{3}+2\sqrt{2}33+22知识点：垂径定理章节：图形的性质 / 圆
如图,内接于,,为的直径,且与的边交于点,若,,则的长是（ ）如图,内接于,,为的直径,且与的边交于点,若,,则的长是（ ）如图,内接于,,为的直径,且与的边交于点,若,,则的长是（ ）