(1)如图①所示,△ABC的外角平分线交于G,试说明∠BGC=90°- ∠A. 说明:根据三角形内角和等于180°,可知∠ABC+∠ACB=180°-∠_____. 根据平角是180°,可知∠ABE+∠ACF=180°×2=360°, 所以∠EBC+∠FCB=360°
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 13:08:34
![(1)如图①所示,△ABC的外角平分线交于G,试说明∠BGC=90°- ∠A. 说明:根据三角形内角和等于180°,可知∠ABC+∠ACB=180°-∠_____. 根据平角是180°,可知∠ABE+∠ACF=180°×2=360°, 所以∠EBC+∠FCB=360°](/uploads/image/z/10716945-33-5.jpg?t=%EF%BC%881%EF%BC%89%E5%A6%82%E5%9B%BE%E2%91%A0%E6%89%80%E7%A4%BA%2C%E2%96%B3ABC%E7%9A%84%E5%A4%96%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8EG%2C%E8%AF%95%E8%AF%B4%E6%98%8E%E2%88%A0BGC%3D90%C2%B0-+%E2%88%A0A%EF%BC%8E+%E8%AF%B4%E6%98%8E%EF%BC%9A%E6%A0%B9%E6%8D%AE%E4%B8%89%E8%A7%92%E5%BD%A2%E5%86%85%E8%A7%92%E5%92%8C%E7%AD%89%E4%BA%8E180%C2%B0%2C%E5%8F%AF%E7%9F%A5%E2%88%A0ABC%2B%E2%88%A0ACB%3D180%C2%B0-%E2%88%A0_____%EF%BC%8E+%E6%A0%B9%E6%8D%AE%E5%B9%B3%E8%A7%92%E6%98%AF180%C2%B0%2C%E5%8F%AF%E7%9F%A5%E2%88%A0ABE%2B%E2%88%A0ACF%3D180%C2%B0%C3%972%3D360%C2%B0%2C+%E6%89%80%E4%BB%A5%E2%88%A0EBC%2B%E2%88%A0FCB%3D360%C2%B0)
(1)如图①所示,△ABC的外角平分线交于G,试说明∠BGC=90°- ∠A. 说明:根据三角形内角和等于180°,可知∠ABC+∠ACB=180°-∠_____. 根据平角是180°,可知∠ABE+∠ACF=180°×2=360°, 所以∠EBC+∠FCB=360°
(1)如图①所示,△ABC的外角平分线交于G,试说明∠BGC=90°- ∠A. 说明:根据三角形内角和等于180°,
可知∠ABC+∠ACB=180°-∠_____. 根据平角是180°,可知∠ABE+∠ACF=180°×2=360°, 所以∠EBC+∠FCB=360°-(∠ABC+∠ACB)=360°-(180°-∠_____)=180°+∠______.根据角平分线的意义,可知∠2+∠3= (∠EBC+∠FCB)= (180°+∠_____)=90°+ ∠_______.所以∠BGC=180°-(∠2+∠3)=90°-∠____. (2)如图②所示,若△ABC的内角平分线交于点I,试说明∠BIC=90°+ ∠A. (3)用(1),(2)的结论,你能说出∠BGC和∠BIC的关系吗
(1)如图①所示,△ABC的外角平分线交于G,试说明∠BGC=90°- ∠A. 说明:根据三角形内角和等于180°,可知∠ABC+∠ACB=180°-∠_____. 根据平角是180°,可知∠ABE+∠ACF=180°×2=360°, 所以∠EBC+∠FCB=360°
(1)A A A A A A (2)说明:根据三角形内角和等于180°,可得∠ABC+∠ACB=180°-∠A,根据角平分线的意义,有 ∠6+∠8= (∠ABC+∠ACB)= (180°-∠A)=90°- ∠A,所以∠BIC=180°-(∠6+∠8) =180°-(90°- ∠A) =90°+ ∠A,即∠BIC=90°+ ∠A. (3)互补.