如图,在△ABC中,三条内角的平分线AD,BE,CF相交于I点,IH⊥BC,求证,∠BID=∠HIC
问题描述:
如图,在△ABC中,三条内角的平分线AD,BE,CF相交于I点,IH⊥BC,求证,∠BID=∠HIC
答
证明:∵AD平分∠BAC∴∠BAI=∠BAC/2∵BE平分∠ABC∴∠CBI=∠ABC/2∵CF平分∠ACB∴∠BCI=∠ACB/2∴∠BID=∠BAI+∠ABE=(∠BAC+∠ABC)/2∵∠BAC+∠ABC=180-∠ACB∴∠BID=(180-∠ACB)/2=90-∠ACB/2=90-∠B...