Question

如图,在三角形ABC中,AC=14,BC=8,D为AB的中点,E为AC边上的一点,且∠AED=90°+1/2∠

∠AED=90°+1/2∠C,求CE的长

Answer 1

取 AC 中点F，连接DF, 则
DF//BC
DF = BC/2 = 8/2 = 4
∠EFD = ∠C
∠AED = 90 + ∠C/2
∠AED = 180 - ∠DEF
∴∠DEF = 90 - ∠C/2
∠AED 是三角形 DEF外角
∠AED = ∠EFD + ∠EDF = ∠C + ∠EDF
∠EDF = ∠AED - ∠C = 90 + ∠C/2 - ∠C = 90 - ∠C/2
∴∠DEF = ∠EDF
三角形 FED 是等腰三角形
FE = FD = BC/2 = 4
∴CE = CF + FE = AC/2 + FE = 14/2 + 4 = 11