a) Ta có: \(\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} = \)\(\overrightarrow {AC} \) + \(\overrightarrow {CD} \) = \(\overrightarrow {AD} \) = \(\overrightarrow {AE} + \overrightarrow {ED} \) = \(\overrightarrow {AE} - \overrightarrow {DE} \).

Vậy \(\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {AE} - \overrightarrow {DE} \).

b) Ta có: \(\overrightarrow {AB} + \overrightarrow {BD} = \overrightarrow {AD} = \overrightarrow {AE} + \overrightarrow {ED} \)

       ⇒ \(\overrightarrow {AB} + \overrightarrow {BD} = \overrightarrow {AE} + \overrightarrow {ED} \)

     ⇔ \(\overrightarrow {AB} - \overrightarrow {ED} = \overrightarrow {AE} - \overrightarrow {BD} \)

     ⇔ \(\overrightarrow {AB} + \overrightarrow {DE} = \overrightarrow {AE} - \overrightarrow {BD} \)

Vậy ta có đpcm.

c) Ta có: \(\overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {BD} = \overrightarrow {BE} + \overrightarrow {ED} \)

           ⇒ \(\overrightarrow {BC} + \overrightarrow {CD} = \overrightarrow {BE} + \overrightarrow {ED} \)

            ⇔ \(\overrightarrow {BC} - \overrightarrow {ED} = \overrightarrow {BE} - \overrightarrow {CD} \)

           ⇔ \(\overrightarrow {BC} + \overrightarrow {DE} = \overrightarrow {BE} - \overrightarrow {CD} \)

Vậy ta có đpcm.