Question

如图,在三角形ABC中,AB=AC,点D. E 分别在BC AC上,且角ADE=角AED,角EDC=20 求角BAD=?

Answer 1

∠AED=∠EDC+∠C,∠ADC=∠B+∠BAD,
∵AD=AE,
∴∠AED=∠ADE,
∵AB=AC,
∴∠B=∠C,
∴∠B+∠BAD=∠EDC+∠C+∠EDC,
即∠BAD=2∠EDC,
∵∠EDC=20°,
∴∠BAD=40°．