如图,已知AO是等腰三角形AEF的底边EF上的高,有AO=EF,延长AE到B,使BE=AE,过点B作AF的垂线,垂足为G
1个回答

延长AO交BC于点D,连接BO;过点E作EG‖BO,交AO于点G.

已知,BE = AE ,EG‖BO,

可得:OG = AG = (1/2)AO = (1/2)EF = OE ;

所以,∠BOD = ∠EGO = ∠GEO = ∠BOE .

已知,AO是等腰△AEF的底边EF上的高,可得:AO平分顶角∠EAF;

∠ADC = 90°-∠CAO = 90°-∠BAO = ∠AEO ,

∠OBD = ∠ADC-∠BOD = ∠AEO-∠BOE = ∠OBE ,

所以,点O是∠ABC的平分线BO和∠BAC的平分线AO的交点,

即有:点O是△ABC的内心.