Question

如图,三角形abc中,ac是圆o的直径,ab交o于d,e为bc的中点,且de切圆o于点d.图和问题补充,

Answer 1

连接OD,OE.

1,∠ADC=∠ADO+∠ODC ∠OED=∠ODC+∠CDE

∠ADC=∠OED=90 ∠ADO=∠CDE

OD=OA ∠A=∠ADO

∠A=∠CDE

2,E为CB中点,OE为三角形ABC中位线

OE//AB ∠COE=∠A ∠DOE=∠ODA

∠COE=∠DOE OD=OC,OE=OE

△COE≌△DOE ∠OCE=∠OED=90

BC是O的切线

3,CD^2=AC^2-AD^2

CD=√5 CB^2=CD^2+BD^2=5+BD^2

CB^2=AB^2-AC^2

5+BD^2=(AD+BD)^2-9

BD=5/2