RT
证明:∠CGH=90°-1/2∠ACB=90°-1/2[180°-(∠ABC+∠CAB)]=1/2(∠ABC+∠CAB)=1/2∠ABC+1/2∠CAB=∠ABG+∠BAG又∵∠ABG+∠BAG=∠BGD∴∠CGH=∠BGD得证!