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.
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.
数学人气:402 ℃时间:2020-04-08 05:11:08
优质解答
∵BD平分∠ABC,CE平分∠ACE,
∴∠DBC=
∠ABC,∠ECB=
∠ACB,
∴∠BIC=180°-(∠DBC+∠ECB)
=180°-(
∠ABC+
∠ACB)
=180°-
(∠ABC+∠ACB)
=180°-
(180°-∠A)
=90°+
∠A.
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