解:(1)∵BD、CD是∠ABC和∠ACB的角平分线, ∴∠DBC=∠ABC,∠DCB=∠ACB, ∵∠ABC+∠ACB=180°-∠A, ∠BDC=180°-∠DBC-∠DCB=180°-(∠ABC+∠ACB)=180°-(180°-∠A)=90°+∠A, ∴∠BDC=90°+∠A。 (2)∵BD、CD是∠ABC和∠ACB外角的平分线, ∴∠CBD=(∠A+∠ACB),∠BCD=(∠A+∠ABC), ∵∠ABC+∠ACB=180°-∠A, ∠BDC=180°-∠CBD-∠BCD=180°-(∠A+∠ACB+∠A+∠ABC)=180°-(2∠A+180°-∠A)=90°-∠A 即∠BDC=90 °-∠A。 |