✔ 最佳答案
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Construct a point G on AC, such that making a(GBE) be 30, thus a(GBC) be 30 30 =60, as a(BFC) be 60 it forms a equal tri(FBC). We get FB=FC=BC.
As a(GBE)=a(EBC)=30 , EB bisects FC, We get two same tri(EHF) and tri(EHC), so a(HEC)=180-20-90=70=a(HEF), and a(EFC)=a(ECF)=20
a(BDC)=180-30-50-60=40. for FB=FC , a(BFD)=180-20-40=120=a(CFG), a(FBD)=20=a(FCG), a(BDC)=a(CGB)=40, tri(DFB)=tri(GFC), so DB=GC.
Furthermore, a(DFG)=a(BFC)=60 and DF=FG, so a(FDG)=a(FGD)=(180-60)/2=60, so tri(DFG) is an equal tri, so DF=FG
a(HEC)=180-90-20=70, a(HEF)=70(proved above), so a(FEG)=180-70-70=40.
a(FGE)=180-30-30-60-20=40, a(FEG)=a(FGE), so FE=FG. we get DF=FG and FE=FG, so DF=FE.
We have got a(EFC)=20, a(CFB)=60, a(BFD)=120, so a(DFE)=360-20-60-120=160.
As DF=FE, a(DEF)=(180-160)/2=10. x=a(DEF) a(HEF)=10+70=80
有D亂, 唔好意思
