由于∠ECD=∠ACB=90,所以∠ECA=∠DCB,又EC=DC,AC=BC,故△ECA全等于△DCB,从而AE=BD,∠BDC=∠AEC=45,推出∠BDE=∠EDC+∠BDC=90.故知四边形ACBD是正方形,AD=BD=AE=AC,得证