两道八下几何证明题!不是很难,只需要列算式!
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两道八下几何证明题!不是很难,只需要列算式!
1.(1)如图2,求证∠A+∠DBE+∠C+∠D+∠E=180°
(2)如图3,求证∠A+∠DBE+∠C+∠D+∠E=180°
2.如图1,求证∠A+∠B+∠C+∠D+∠E+∠F+∠G=180°
![](http://img.wesiedu.com/upload/f/fa/ffaa63c87d7d4b35528e1affb6dbb91b.jpg)
1.(1)如图2,求证∠A+∠DBE+∠C+∠D+∠E=180°
(2)如图3,求证∠A+∠DBE+∠C+∠D+∠E=180°
2.如图1,求证∠A+∠B+∠C+∠D+∠E+∠F+∠G=180°
![](http://img.wesiedu.com/upload/f/fa/ffaa63c87d7d4b35528e1affb6dbb91b.jpg)
![两道八下几何证明题!不是很难,只需要列算式!](/uploads/image/z/7464390-6-0.jpg?t=%E4%B8%A4%E9%81%93%E5%85%AB%E4%B8%8B%E5%87%A0%E4%BD%95%E8%AF%81%E6%98%8E%E9%A2%98%21%E4%B8%8D%E6%98%AF%E5%BE%88%E9%9A%BE%2C%E5%8F%AA%E9%9C%80%E8%A6%81%E5%88%97%E7%AE%97%E5%BC%8F%21)
1,(1)设CE交BD与F,交AD于G,
∠B+∠E+∠BFE
=∠B+∠E+∠D+∠CGD
=∠B+∠E+∠D+∠A+∠C=180
(2)∠CBD+∠DBE+∠ABE
=∠A+∠D+∠DBE+∠C+∠E=180
2,设BD交CF于N,交CE于M;CF交AG于X,交AD于H
∠B+∠E+∠BME
=∠B+∠E+∠C+CNM
=∠B+∠E+∠C+∠D+∠CHD
=∠B+∠E+∠C+∠D+∠A+∠AXH
=∠B+∠E+∠C+∠D+∠A+∠G+∠F=180
∠B+∠E+∠BFE
=∠B+∠E+∠D+∠CGD
=∠B+∠E+∠D+∠A+∠C=180
(2)∠CBD+∠DBE+∠ABE
=∠A+∠D+∠DBE+∠C+∠E=180
2,设BD交CF于N,交CE于M;CF交AG于X,交AD于H
∠B+∠E+∠BME
=∠B+∠E+∠C+CNM
=∠B+∠E+∠C+∠D+∠CHD
=∠B+∠E+∠C+∠D+∠A+∠AXH
=∠B+∠E+∠C+∠D+∠A+∠G+∠F=180