如题所述
根据三角形外角定理:
∠1=∠2+∠D,→∠D=∠1-∠2,
∠ACE=∠ABC+∠A,
∵BD、CD是角平分线,
∴∠ACE=2∠1,∠ABC=2∠2,
∴2∠1=2∠2+∠A,
∠A=2(∠1-∠2)=2∠D,
∵BD、CD分别平分∠ABC、∠ACE,∴∠2=1/2∠ABC,∠1=1/2∠ACE,∵∠1=∠2+∠D,∠ACE=∠ABC+∠A,∴∠ACE-∠ABC=∠A,∴∠D=∠1-∠2=1/2(∠ACE-∠ABC),=1/2∠A。向左转|向右转