Question

如图所示,在三角形ABC中,AB=AC,AD=AE,角BAD=60度,求角EDC的度数

Answer 1

∵AB=AC

∴∠B=∠C=(180°-∠BAC)/2

∵AD=AE

∴∠ADE=∠AED=(180°-∠DAE)/2

∵∠ADC=∠ADE+∠EDC=∠B+∠BAD

∴∠EDC=∠B+∠BAD-∠ADE=(180°-∠BAC)/2+∠BAD-(180°-∠DAE)/2

=∠BAD-(1/2)(∠BAC-∠DAE)=∠BAD-(1/2)∠BAD=(1/2)∠BAD=(1/2)×60°=30°

即∠EDC=30°