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?
答
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.