In the given figure, EAEC = EB ED , prove that DEAB ~ DECD

In \(\displaystyle \triangle \mathrm{EAB}\) and \(\displaystyle \triangle \mathrm{ECD}\) \[\begin{aligned} & \frac{\mathrm{EA}}{\mathrm{EC}}=\frac{\mathrm{EB}}{\mathrm{ED}} \\ & \angle \mathrm{AEB}=\angle \mathrm{CED} \\ & \angle \mathrm{EAB} \sim \triangle \mathrm{ECD} \end{aligned} \]