A) \[CD=DE\]
B) \[ED\] is not parallel to\[AB\]
C) ED passes through the centre of the circle
D) a CDE is a right angled triangle
Correct Answer: A
Solution :
Join\[ED\], then \[\angle DEC=\angle ACD={{50}^{o}}\] (angles in alternate segment) \[\angle EDC=\angle BCE={{50}^{o}}\] (cyclic in alternate segment) \[\therefore \] \[\angle DEC=\angle EDC\] So, \[CD=CE\]You need to login to perform this action.
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