今早那能:
证明:
∵BD、CD分别是∠ABC、∠ACB的角平分线
∴∠DBC=1/2∠ABC,∠DCB=1/2∠ACB
在△ABC中,∠ABC+∠ACB=180°-∠A
在△BCD中,
∠BDC=180°-(∠DBC+∠DCB)
=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
今早那能:
证明:
∵BD、CD分别是∠ABC、∠ACB的角平分线
∴∠DBC=1/2∠ABC,∠DCB=1/2∠ACB
在△ABC中,∠ABC+∠ACB=180°-∠A
在△BCD中,
∠BDC=180°-(∠DBC+∠DCB)
=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A