question_answer

\[ACB\] is a tangent to a circle at\[C\]. \[CD\] and \[CE\] are chords such that \[\angle ACE>\angle ACD\]. If\[\angle ACD=\angle BCE={{50}^{o}}\], then:

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,       \[CD=CE\]