三角形ABC中,三条内角平分线AD,BE,CF相交于点G,GH垂直于BC于H点,求证：角BGD=角HGC.

Question

三角形ABC中,三条内角平分线AD,BE,CF相交于点G,GH垂直于BC于H点,求证：角BGD=角
HGC.

Answer 1

证明：∵∠AEG=∠EBC+∠ACB= 1/2∠ABC+∠ACB,
∴∠AGE=180°-（∠DAC+∠AEG）
=180°-[1/2∠BAC+1/2∠ABC+∠ACB]
=180°-[1/2（∠BAC+∠ABC）+∠ACB]
=180°-[1/2 （180°-∠ACB）+∠ACB]
=180°-[90°+1/2∠ACB]
=90°-1/2∠ACB,
∴∠BGD=∠AGH=90°-1/2∠ACB,
又∵在直角△GCH中,∠CGH=90°-∠GCD=90°-1/2 ∠ACB,
∴∠BGD=∠CGH．
希望采纳