题目内容
如图,AB∥CD,直线l分别与AB、CD相交,若∠1=130°,则∠2=
A.40°
B.50°
C.130°
D.140°
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.