题目如图,ABABAB是半圆OOO的直径,CD^=2BD^\widehat {CD}=2\widehat {BD}CD=2BD,CDCDCD的延长线交ABABAB的延长线于点EEE.若∠E=40∘\angle E=40^{\circ}∠E=40∘,则CD^\widehat {CD}CD的度数为______∘^{\circ}∘;若CDDE=65\frac{CD}{DE}=\frac{6}{5}DECD=56,则BEAB=______.\frac{BE}{AB}= \_\_\_\_\_\_.ABBE=______.知识点:切线的性质章节:图形的性质 / 圆