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Eight problems of three-year simulated mathematics in five-year senior high school entrance examination
The extension line across d, DF, vertical CB and CB is at point F.

From the known AC=BC

Angle CAB= angle CBA

AD=BD

Angle DAB= angle DBA

So angle DAE= angle DBF

Angle DEA= Angle DFB = 90°.

AD=BD (corner edge)

So DE=DF

Because angle C= angle E= angle F= angle EDF.

CEDF is square.

DE=CE=AE+AC=AE+BC

So ED=AE+BC

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