Question

ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:

(i) AD || BC

(ii) EB = EC.

Answer 1

(i) If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA = ED, then we have to prove the following, AD || BC

(ii) EB = EC

(i) It is given that EA = ED, so

$\angle EAD=\angle EDA=x$

Since, ABCD is cyclic quadrilateral

Now,

Therefore, the adjacent angles andare supplementary

Hence, AD || BC

(ii) Since, AD and BC are parallel to each other, so,

$$\begin{gathered} \angle ECB=\angle EDA\left(\mathrm{Corresponding}\mathrm{angles}\right) \\ \angle EBC=\angle EAD\left(\mathrm{Corresponding}\mathrm{angles}\right) \\ \mathrm{But},\angle EDA=\angle EAD \\ \mathrm{Therefore},\angle ECB=\angle EBC \\ \Rightarrow EC=EB \\ \mathrm{Therefore},△ECB\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}. \end{gathered}$$