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Question
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
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Solution
EB = EC
Since, AD and BC are parallel to each other, so,
\[\angle ECB = \angle EDA \left( \text{ Corresponding angles } \right)\]
\[\angle EBC = \angle EAD \left( \text{ Corresponding angles} \right)\]
\[\text{ But } , \angle EDA = \angle EAD\]
\[\text{ Therefore } , \angle ECB = \angle EBC\]
\[ \Rightarrow EC = EB\]
\[ \text{ Therefore, } \bigtriangleup \text{ ECB is an isosceles triangle } .\]
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