Question

如图,在△ABC中,∠B=∠C ,∠ADE=∠AED.求证：∠EDC=二分之一∠BAD

Answer 1

①∠ADC=∠ADE+∠EDC②∠ADC=∠BAD+∠B推出∠ADE+∠EDC=∠BAD+∠B已知∠ADE=∠AED,∴∠AED+∠EDC=∠BAD+∠B∵∠AED=∠C+∠EDC带入上式,得∠C+∠EDC+∠EDC=∠BAD+∠B已知∠B=∠C ,∴2∠EDC=∠BAD,即∠EDC=二分之一∠B...