1.作CD垂直AB于D,则∠ACD=30°,AC=2AD.
设AD=X,则AC=2X,CD=√3X;又∠B=45°,则BD=CD=√3X.
AB=3+√3=X+√3X,X=3,CD=3√3.
故:S⊿ABC=AB*CD/2=(3+√3)*(3√3)/2=(9√3+9)/2.
2.证明:(1)连接DF.CD为直径,则:∠DFC=90°,DF⊥BC;
又AC⊥BC.故DF∥AC,BF/FC=BD/DA.
∵BD=DA.
∴BF=FC,即点F为BC中点.
(2)∵DF∥AC.
∴∠CEF=∠DFE;
又∠DGE=∠DFE.
∴∠DGE=∠CEF.(等量代换)
又∠GEC=∠DGE+∠A,即∠GEF+∠CEF=∠DGE+∠A.
∴∠A=∠GEF.