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∠

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