等腰三角形ABC ∠B=∠C,点D在BC边上,点E在AC上,并且∠ADE=∠AED,∠BAD=20,求EDC∠的度数?

问题描述:

等腰三角形ABC ∠B=∠C,点D在BC边上,点E在AC上,并且∠ADE=∠AED,∠BAD=20,求EDC∠的度数?

∵∠B=∠C
∴180°-∠A=2∠C (1)
又∵∠DAE=∠A-20°
∴180°-∠DAE=180°-(∠A-20°)=200°-∠A=2∠DAE (2)
(2)-(1)
得:20°=2∠DAE-2∠C
即:∠DAE-∠C =10°,∠DAE=∠C +10°,
∵∠DAE=∠C +∠EDC
∴∠EDC=10°

10度。20+B=ADE+EDC,而ADE=AED=EDC+C带入得10

ADC=ADE+CDE,又ADE=AED=C+CDE,所以ADC=C+2CDE.
又ADC=B+BAD,B=C,所以BAD=2CDE.
BAD=20,所以CDE=10.