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