如图9,直线DF雨GE相交于点B,BD平分∠ABC,∠GBF=21°,∠CBE=2∠ABE,求∠ABC的度数
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![如图9,直线DF雨GE相交于点B,BD平分∠ABC,∠GBF=21°,∠CBE=2∠ABE,求∠ABC的度数](/uploads/image/z/14563581-69-1.jpg?t=%E5%A6%82%E5%9B%BE9%2C%E7%9B%B4%E7%BA%BFDF%E9%9B%A8GE%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9B%2CBD%E5%B9%B3%E5%88%86%E2%88%A0ABC%2C%E2%88%A0GBF%3D21%C2%B0%2C%E2%88%A0CBE%3D2%E2%88%A0ABE%2C%E6%B1%82%E2%88%A0ABC%E7%9A%84%E5%BA%A6%E6%95%B0)
如图9,直线DF雨GE相交于点B,BD平分∠ABC,∠GBF=21°,∠CBE=2∠ABE,求∠ABC的度数
如图9,直线DF雨GE相交于点B,BD平分∠ABC,∠GBF=21°,∠CBE=2∠ABE,求∠ABC的度数
如图9,直线DF雨GE相交于点B,BD平分∠ABC,∠GBF=21°,∠CBE=2∠ABE,求∠ABC的度数
∵∠GBF=21°
∴∠DBE=21°
∵∠CBE=2∠ABE
∴∠ABC=∠CBE+∠ABE=2∠ABE+∠ABE=3∠ABE
∵BD平分∠ABC
∴∠DBA=∠ABC/2=3∠ABE/2=1.5∠ABE
∴∠DBE=∠DBA-∠ABE=1.5∠ABE-∠ABE=0.5∠ABE
∵∠DBE=21°
∴∠ABE=∠DBE*2=21°*2=42°
∠ABC=3∠ABE=3*42°=126°
易证:∠CBD=∠DBE=∠ABE
对顶角:∠FBE=DBE=21
∠ABE=3∠DBE=63
∵BD平分∠ABC,∠GBF=21°,∠CBE=2∠ABE,
∴∠CBD=∠DBA,∠DBE=∠GBF=21°
∴∠CBE-∠DBE=∠DBE+∠EBA
∴2∠ABE-21°=∠ABE+21°
∴∠ABE=42°
∴∠ABC=2(∠ABE+∠DBE)=126°
∵∠CBD=∠ABD=(1/2)∠ABC.
∠DBE=GBF-21°
∠CBE=2∠ABE.
∠ABC=∠CBE+∠ABE=3∠ABE,
∠ABE=∠ABD-21°
=(1/2)∠ABC-21° .
∠ABC=3∠ABE.
=3([(1/2)∠ABC-21°].
=(3/2)∠ABC-63°
(3/2-1)∠ABC=63°.
∴∠ABC=126° .