证明：
(1) $\because △ ACB$ 和$△ ECD$ 都是等腰直角三角形,
$\therefore AC=BC,EC=DC.$
$\because ∠ ACE=∠ DCE-∠ DCA,∠ BCD=∠ ACB-∠ DCA,∠ ACB=∠ ECD=90°$,
$\therefore ∠ ACE=∠ BCD.$
$\therefore △ ACE≌△ BCD(\mathrm{SAS}).$
(2) $\because △ ACE≌△ BCD$,
$\therefore ∠ EAC=∠ DBC,AE=BD.$
$\because ∠ DBC+∠ DAC=90°$,
$\therefore ∠ EAC+∠ DAC=∠ EAD=90°.$
$\therefore AD^2+AE^2=DE^2.$
$\because ∠ DCE=90°,CD=CE$,
$\therefore CD^2+CE^2=DE^2$,
$\therefore 2CD^2=DE^2.$
$\therefore AD^2+AE^2=2CD^2.$
$\because AE=BD$,
$\therefore AD^2+BD^2=2CD^2.$
