现有三角形ABC, 点D, E 位于边CA, AB上, BD等分角B, CE等分角C.
已知角BDE=24度, 角CED=18度.
求三角形ABC的三个角.
解.

CE和BD的交点记为O, 那么在△DOE中:
∠DOE = 180 - 18 - 24 = 138
∠BOC = ∠DOE = 180 - (∠B + ∠C)/2 = 138

∠B + ∠C = 84
∠A = 180 - ∠B - ∠C = 96

在BC上取一点F, 使BF=BE, 取另一点G, 使CG=CD.

连接EF, EF与BD的交点记为L, 因为EL=FL,△DEL和△DFL互为镜象,
∠BDF = ∠BDE = 24

连接DG, DG与CE的交点记为M, 因为DM=GM,△DEM和△GEM互为镜象,
∠CEG = ∠CED = 18

连接EG, EG与BD的交点记为N,
∠BNE = ∠BNF = ∠DEG + ∠BDE = 2*18+24 = 60
∠FNG = 180 - ∠BNE - ∠BNF = 60
所以, NF平分∠BNG.

∠EGD = ∠EDG = 90 - ∠CED = 72
∠FDG = ∠EDG - ∠BDE - ∠BDF = 72–24–24 = 24
所以, DF平分∠BDG.

对于△DGN, 点F与两个顶点的连线平分一个内角和一个外角, 点F必定是△DGN的一个旁心, GF也必定平分∠DGN的外角, 所以
∠BGE = ∠CGD

又 ∠BGE + ∠CGD + ∠EGD = 180
所以∠CGD = (180 - ∠EGD)/2 = (180–72)/2 = 54

在直角三角形△CGM中, ∠C/2 + ∠CGD = 90, 所以
∠C = 2 * (90–54) = 72
∠B = 84 - ∠C = 12