∫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