Question

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?

Answer 1

解：∠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°

Answer 2

∠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°

Answer 3

∵∠ADE=∠AED=∠C+20°
∴∠ADC=∠C+40°
∵∠ADC=∠B+∠BAD
∠B=∠C
∴∠BAD=40°

Answer 4

∠BAD=∠ADC-∠B
=∠ADC-∠C
=∠ADE+∠EDC-∠C
=∠AED+∠EDC-∠C
=∠AED+20-∠C
=(∠AED-∠C) +20
=40

Answer 5

设∠ADE=∠AED=x，∠BAD=y，∠B=x+20°-y，∠C=x-20°所以x+20°-y=x-20°，y=40°=∠BAD

Answer 6