Question

正方形ABCD,AF切以BC为直径的半圆与E,交CD于F,若AB=4.求:CF

抱歉 图我弄不上

Answer 1

设O是以BC为直径的半圆的圆心
连接OF，OA
AE切圆O
所以OE⊥AF
∴△OEF≌△OCF，△AEO≌△ABO
∴CF=EF，AE=AB
∵∠AOE=∠AOB，∠EOF=∠COF
∴∠AOF = 90°
所以
OE" = AE×EF
2OE =BC
∴OE = 2
AE = AB = 4
∴EF = 1
∴CF = 1