如图所示,在△ABC中,AC=7,BC=4,D为AB的中点,E为AC边上一点,且∠AED=90°+（1/2）∠C

过D做DP平行BC，交AC于P
∵∠AED=90°+（1/2）∠C，（1/2）∠C<90°
∴∠AED=90°+（1/2）∠C>∠C
∴p点在CE上
∵D为AB的中点
∴P为AC中点
∴DP=BC/2=2（中位线），PC=AC/2=7/2
∴∠EPD=∠C
∴∠EDP=∠AED-∠EPD
=90°+1/2∠C-∠C=90°-1/2∠C
∵∠PED=180°-∠AED
=180°-90°-1/2∠C=90°-1/2∠C
∴∠PED=∠PDE
∴EP=DP=2
∴CE=EP+PC=2+7/2=11/2

取 AC 中点F，连接DF, 则
DF || BC
DF = BC/2 = 4/2 = 2
∠EFD = ∠C

∠AED = 90 + ∠C/2
∠AED = 180 - ∠DEF
所以
∠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

因此
∠DEF = ∠EDF
三角形 FED 是等腰三角形
FE = FD = BC/2 = 2

所以
CE = CF + FE = AC/2 + FE = 7/2 + 2 = 11/2