∫AEF =∠bec ∴∠bec=45

∫≈ B = 90 ∴ Be = BC ∫ BC = 3 ∴ Be = 3.

(2) the intersection e is EG⊥CN, and the vertical foot is g point.

∴be=cg∵ab∨cn ∴∠aeh=∠n,∠bec=∠ecn

∵∠AEH=∠BEC∴∠N=∠ECN ∴EN=EC

∴CN=2CG 2BE

bw = x，DN=Y，CD=AB=4 ∴Y=2X-4(2≤X≤3)

(3) ∠ bad =90 ∴∠AEF+∠AFE=90

∵EF⊥EC ∴∠AEF+∠CEB=90

∴∠AFE=∠CEB ∴∠HFE=∠AEC

When △FHE is similar to △AEC,

FHE =∠ East African Community

∠∠bad =∠b，∠AEH=∠BEC ∴∠FHE=∠ECB ∴∠EAC=∠ECB

∴tan∠eac=tan∠ecb ∴bc:ab=be:bc ∴be=9/4 ∴dn=？

If ∠ fhe = ∠ ECA, let EG⊥DC be in G and AC be in O.

* en = EC EG⊥CN ∴∠NEG=∠GEC.

∵ Ah ∨ For example ∴∠FHE=∠NEG ∴∠FHE=∠GEC.

∴∠GEC=∠ECA ∴EO=OC

Let EO=C0=3K and AE=4K AO=5K.

AO+CO=8K=5 ∴K=？

∴AE=5/2,BE=3/2

∴DN= 1