(1), according to the equilateral right angle △ theorem: AD=CD=BD.

Because △EFB and △GDB are right triangles, ∠DGB=∠CED,

∠ EDC =∠ GDB = 90 degrees, DC = decibel.

So △ EDC △ GDB = > ED=GD

Because AD=CD and AE=CG.

(2) Because BE=BD+DE=CD+DE, the line segment equal to it may be CM.

That is, it is necessary to prove that DM=DE.

Because ∠AEH=∠CED, ∠ AHE = ∠ CDE = 90.

So ∠ DCE = ∠ EAH = > ∠DCE =∠ dam

At △ADM and △CDE:

∠ADM=∠CDE=90

∠DCE =∠ dam

AD=CD

So △ ADM△ CDE

So DM=DE

So CM=BE