chenbotong
幼苗
共回答了12个问题采纳率:100% 举报
1,已知角DAE=角BAC,∠FEA=1/2∠DEA,∠FCA=1/2∠BCA,∠DEA=180-∠DAE-∠D,角BCA=180-∠BAC-角B 连接FA并延长,则 而∠FEA+∠FCA=1/2(∠DEA+∠BCA)=1/2(180-∠DAE-∠D+180-∠BAC-角B )又角DAE=角BAC,所以原式=180-∠BAC-(∠B+∠D)/2 ∠EAC=角F+∠FEA+∠FCA=∠F+180-∠BAC-(∠B+∠D)/2 ∠EAC=∠F+180-∠BAC-(∠B+∠D)/2 移项,得 ∠EAC +∠BAC =∠F+180-∠BAC-(∠B+∠D)/2 180=∠F+180-∠BAC-(∠B+∠D)/2 ∠F=(∠B+∠D)/2 2,将2,4代人,得a=3
1年前
10