Question

如图,AB是圆O的直径,AC与圆O相切于点A,且AC等于AB等于2,连接OC交圆O于点D,BD的延长线交AC于E 求AE

Answer 1

连接AD,设AE=x,则CE=2-x.∵∠DAC+∠DAB=∠OED+∠DAB=90°∠OED=∠OBD∴∠DAC=∠OBD=∠CDE又∵∠DCE=∠ACD∴△DCE∽△ACD∴CE/CD=CD/CA又∵AC=AB=2,∴CD=√(4-2x)OA=1,OC=√5OC=OD+CD∴√5=1+√(4-2x)解得AE=x=-1+√5...