如图在三角形ABC中,∠B=∠C,点D在BC边上,∠BAD=50°,在AC上取一点E使∠ADE=∠AED,则∠EDC的度数为

Question

Answer 1

∵∠ADC＝∠BAD+∠B,∠ADC＝∠ADE+∠EDC
∴∠BAD+∠B＝∠ADE+∠EDC
∴∠ADE＝∠BAD+∠B-∠EDC
∵∠ADE＝∠AED
∵∠AED＝∠EDC +∠C
∴∠BAD+∠B-∠EDC＝∠EDC +∠C
∴2∠EDC＝∠BAD+∠B-∠C
∵∠B＝∠C
∴2∠EDC＝∠BAD
∵∠BAD＝50°
∴∠EDC＝25°