如图,三角形ABC,角C=90度,角ABC的角平分线与角BAC的外角平分线相交于点E,求证:角E=45度
人气:413 ℃ 时间:2019-08-20 22:56:28
解答
∠EAC=1/2(∠C+∠ABC)
∠E=180°-∠EAB-∠ABE
=180°-(∠EAC+∠BAC)-∠ABE
=180°-1/2∠C-1/2∠ABC-∠BAC-∠ABE
=180°-45°-(∠BAC+1/2∠ABC+∠ABE)
=135°-(∠BCA+∠ABC)
=135°-90°
=45°
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