△ABC ∠ABC＝∠ACB,D为BC上一点,E为直线AC上一点,∠ADE＝∠AED求证∠BAD＝2∠CDE

设∠CDE=∠1,∠ADE=∠2,∠AED=∠3,∠BAD=∠4.
∠1=∠3-∠ACB
∠4=∠2+∠1-∠ABC∠2
∠4=∠2+∠3-∠ACB-∠ABC
∵∠2=∠3,∠ABC=∠ACB
∴∠4=∠3+∠3-∠ACB-∠ACB
∠4=2∠3-2∠ACB
∠4=2(∠3-∠ACB)
∠4=2∠1
即：∠BAD=2∠CDE