已知AB=AD,BC=DE,∠BAD=∠EDC,求证AC=AE,∠CAE=∠CDE

已知AB=AD,BC=DE,∠BAD=∠EDC,求证AC=AE,∠CAE=∠CDE
已知AB=AD,BC=DE,∠BAD=∠EDC,求证AC=AE,∠CAE=∠CDE

已知AB=AD,BC=DE,∠BAD=∠EDC,求证AC=AE,∠CAE=∠CDE
证明:(1)AB=AD,则:∠ABD=∠ADB;又∠BAD=∠EDC.
故:180°-∠BAD-∠ADB=180°-∠EDC-∠ADB,即∠B=∠ADE;
又AB=AD;BC=DE.故⊿ABC≌ΔADE(SAS),得:AC=AE.
(2)⊿ABC≌ΔADE(已证),则:∠BAC=∠DAE.
故:∠BAC-∠DAC=∠DAE-∠DAC,即∠BAD=∠CAE;
又∠BAD=∠CDE.所以:∠CAE=∠CDE.(等量代换)

答：当∠BAD=2∠CDE时，AD=AE。证明：∵∠AED=∠C ∠EDC ①∠ADE=∠即∠ADE=∠B ∠EDC ②∵ AB=AC ∴∠B=∠C对比①②两式即得到： ∠