1≈a = 20 ,ab=ac,∴∠abc=∠acb=80

∠EBC=60，∠DCB=50 ,∴∠ABE=20，∠ACD=30

In δ δ△BDC.

∠BDC= 180 -∠ABC-∠DCB

= 180 -80 -50

=50 =∠DCB

∴BC=BE

At △BEC

Bec =180-ECB -∠EBC

= 180 -80 -60

=40

If b is BF = BC and BF intersects AC at F, then △ BF=BC is an isosceles triangle.

∴BF=BC=BD

∠ CBF = 180-2 ∠ ACB = 20，∴ FBD = 80-20 = 60。

△ BDF is an equilateral triangle, △ BF = DF.

In △BFE, ∠ FBE = ∠ ABC-∠ Abe -∠ CBF = 80-20-20 = 40 = ∠ FEB.

So ef = BF = df,

△ def is an isosceles triangle.

From ∠ dfe =180-∠ BFC-∠ BFD =180-80-60 = 40.

∠fed = 1/2( 180-∠dfe)= 70。

∴∠DEB=∠FED-BEC=70 -40 =30

2. Proof: Let ∠ HCD = 10, cross DE in G, cross BE in F, and connect DF.

AB = AC，

∴∠ABC=∠ACB，

∫∠A = 20，

∴∠ABC=∠ACB=( 180 -20 )/2=80，

∫∠BCD = 50

∫∠HCD = 10

∴∠HCB=60

∠∠FBC = 60

∴△BCF is an equilateral triangle

∴BC=BF

∫∠BCD = 50

∠∠DBC = 80

∠∠DBC+∠BCD+∠BDC = 180

∴∠BDC=50

∫∠BCD = 50

∴∠BDC=∠BCD

∴BD=BC

∴BD=BF

∴∠BDF=∠BFD

* DBF = 80-* FBC(60)= 20

∴∠BDF=80

∫∠BDC = 50

∴∠CDF=30

∴∠dfh=∠cdf(30)+∠fcd( 10)= 40

∠∠DHF+∠DFH(40)=∠BDF(80)

∴∠DHF=40

∫∠DFH = 40

∴∠DHF=∠DFH

∴DH=DF

BC = BC

∠∠ABC =∠ACB

∠∠HCB =∠EBC

∴△HBC≌△ECB

∴HC=EB

BF = CF

∴HF=EF

∠∠HFE =∠BFC = 60

∴△HFE is an equilateral triangle

∴HE=FE

DH = DF (authentication)

DE = DE

∴△DHE≌△DFE

∴∠HDE=∠FDE

∠DHF(40)+∠FHE(60)+∠HEF(60)+∠EFH(60)+∠HFD(40)+∠HDE+∠FDE = 360

∴∠EDF=50

∫∠CDF = 30

∴∠EDC=80

∴∠DEB=50 +60 -80 =30