Question

如图,三角形ABC中,内角角A和外角CBE和角BCF的角平分线交于点P,AP交BC于D,过B作BG⊥AP于G

Answer 1

1)∠GBP=45°,∠BGP=90°,所以∠BPG=∠GBP=45°,

BP平分∠CBE,所以∠EBP=∠CBP,

∠EBP=∠BAP+∠APB,∠CBP=∠GBP+∠DBG,由于∠APB=∠GBP,所以∠BAP=∠DBG.

△BGD与△ACD中,∠BAP=∠DBG,∠BDG=∠ADC,所以∠ACD=∠BGD=90°,即AC⊥BC.

2)∠APB=∠EBP-∠BAP=∠EBC/2-∠BAC/2=(180°-∠ABC)/2-∠BAC/2=(180°-∠ABC-∠BAC)/2=∠ACB/2；

∠HDC=90°-∠DCP=180°/2-∠BCF/2=180°/2-(180°-∠ACB)/2=∠ACB/2,

所以∠APB=∠HDC