已知:如图,BD平分∠ABC,CE平分∠ACE,BD与CE交于点I,试说明∠BIC=90°+1/2∠A.

问题描述:

已知:如图,BD平分∠ABC,CE平分∠ACE,BD与CE交于点I,试说明∠BIC=90°+

1
2
∠A.

∵BD平分∠ABC,CE平分∠ACE,
∴∠DBC=

1
2
∠ABC,∠ECB=
1
2
∠ACB,
∴∠BIC=180°-(∠DBC+∠ECB)
=180°-(
1
2
∠ABC+
1
2
∠ACB)
=180°-
1
2
(∠ABC+∠ACB)
=180°-
1
2
(180°-∠A)
=90°+
1
2
∠A.