如图,在△ABC中,AB=AC,∠B=∠C,点D,E分别在BC,AC上,且∠ADE=∠AED,∠EDC=20°,则∠BAD的度数为多少?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 22:12:30
![如图,在△ABC中,AB=AC,∠B=∠C,点D,E分别在BC,AC上,且∠ADE=∠AED,∠EDC=20°,则∠BAD的度数为多少?](/uploads/image/z/970966-46-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB%3DAC%2C%E2%88%A0B%3D%E2%88%A0C%2C%E7%82%B9D%2CE%E5%88%86%E5%88%AB%E5%9C%A8BC%2CAC%E4%B8%8A%2C%E4%B8%94%E2%88%A0ADE%3D%E2%88%A0AED%2C%E2%88%A0EDC%3D20%C2%B0%2C%E5%88%99%E2%88%A0BAD%E7%9A%84%E5%BA%A6%E6%95%B0%E4%B8%BA%E5%A4%9A%E5%B0%91%3F)
如图,在△ABC中,AB=AC,∠B=∠C,点D,E分别在BC,AC上,且∠ADE=∠AED,∠EDC=20°,则∠BAD的度数为多少?
如图,在△ABC中,AB=AC,∠B=∠C,点D,E分别在BC,AC上,且∠ADE=∠AED,∠EDC=20°,则∠BAD的度数为多少?
如图,在△ABC中,AB=AC,∠B=∠C,点D,E分别在BC,AC上,且∠ADE=∠AED,∠EDC=20°,则∠BAD的度数为多少?
∠AED=∠C+∠EDC=∠C+20° =∠ADE (1)
又∠ADC=∠B+∠BAD=∠ADE+∠EDC=∠ADE+20° (2)
将(1)代入(2)
∠C+20°+20°= ∠B+ ∠BAD
其中 ∠C= ∠B
∠BAD=40°
∠BAD=∠ADC-∠B
=∠ADC-∠C
=∠ADE+∠EDC-∠C
=∠AED+∠EDC-∠C
=∠AED+20-∠C
=(∠AED-∠C) +20
=40
∵∠ADE=∠AED=∠C+20°
∴∠ADC=∠C+40°
∵∠ADC=∠B+∠BAD
∠B=∠C
∴∠BAD=40°
设∠ADE=∠AED=x,∠BAD=y,∠B=x+20°-y,∠C=x-20°所以x+20°-y=x-20°,y=40°=∠BAD
∠AED=∠EDC+∠C
∠ADC=∠ADE+∠EDC
因为∠ADE=∠AED
所以∠ADC=∠AED+∠EDC
又因为∠BAD=∠ADC-∠B
=∠ADE+∠EDC-∠C
=∠ADE+∠EDC-∠AED+∠EDC
=2∠EDC
=40°