Let $M$ be center of small circle touches the other three sides. $N$ is on ray $ED$ such that $EB=EN$ _If $M\in [EN]$ then we have $\angle BNE=\frac{180^0-\angle E}{2}=\frac{\angle BCD}{2}=\angle BCM$ Hence $BCNM$ is cyclic so $\angle CND=\angle CBM=\frac{180^0-\angle D}{2}\Rightarrow CN=CD$ Thus $EB+CD = ED$ If $N\in [EM]$ similarity we get $EB+CD = ED$ [Moderator edit: Edited improper LaTeX]

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