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°+
∠A.1 2
答
∵BD平分∠ABC,CE平分∠ACE,
∴∠DBC=
∠ABC,∠ECB=1 2
∠ACB,1 2
∴∠BIC=180°-(∠DBC+∠ECB)
=180°-(
∠ABC+1 2
∠ACB)1 2
=180°-
(∠ABC+∠ACB)1 2
=180°-
(180°-∠A)1 2
=90°+
∠A.1 2