以知△ABC中,I是△ABC角平分线的交点,IG⊥BC于G,且∠BAC的角平分线AD交BC于D,求证：∠BID=∠CIG

∠ADB=∠DAC+∠C=1/2∠A+∠C
∠BID=180-∠ADB-∠IBD=180-1/2∠A-∠C-IBD
∵∠IBD=1/2∠B
∴∠BID=180-1/2∠A-∠C-1/2∠B
=180-1/2(∠A+∠B)-∠C
=180-1/2(180-∠C)-∠C
=180-90+1/2∠C-∠C
=90-1/2∠C
∵IG⊥BC
∴∠CIG=90-∠ICG
∵∠ICG=1/2∠C
∴∠BID=∠CIG