题目内容
如图,AB∥CD,直线EF分别交AB、CD于E、F两点,若∠FEB=110°,则∠EFD等于( )
A.50° B.60° C.70° D.110°
C
如图,AB为⊙O的直甲径,PD切⊙O于点C,交AB的延长线于D,且CO=CD,则∠PCA=
A.60°
B.65°
C.67.5°
D.75°
已知如图,AB是半圆直经,△ACD内接于半⊙O,CE⊥AB于E,延长AD交EC的延长线于F,求证:AC·CD=AD·FC.