推荐回答（5个）
悬赏200分又不想确认满意答案，投机的好方法！不过本人不在意悬赏分。
下面三种解题方法，将字母换过即可。其中第三种解法简单明了。
解法1
解法2
解法3
20°，只需要证明AE² = AD * AC
由正弦定理:
AE = AB * sin80°/sin50°
AD = AB * sin60°/sin40°
AC = AB * sin80°/sin20°
所以要证：
(sin80°/sin50°)² = sin60°/sin40° *sin80°/sin20°
即sin20° *sin40° * sin80° = sin²30° * sin60° = √3/8
这是本来是一个恒等式，可以推广，这里形式较简单，可以用三角函数直接证：
sin20° *sin40° * sin80°
= 1/2 (cos20° - cos60°)cos10°
= 1/4[cos30° + cos10° - cos70° - cos50°)
而cos70° + cos50° = 2cos60° * cos10° = cos10°
所以上面等于1/4 cos30° = √3/8，证毕
用三角函数虽然不算得好方法，不过在不知道答案的时候总要把答案猜一下并进行验证。从图上看是很像20°，只是验证一下刚好就是罢了，因为图很简单，用三角函数很合适，就算不是20°也能算出来。
肯定有纯几何的做法，应该要添很多辅助线，以前初中的时候见过几道这样的题，都是图很简单解答很复杂。
用图形的旋转，构成正三角形来解
解：作角FCB=60度，使FC=BC，连接BF,EF,FD以EF为边向下作角EFM=60度，使FM=EF，连接EM,AM
所以三角形FCB是正三角形
所以FC=BC=FB
角BFC=角FCB=60度
三角形EFM是正三角形
所以FM=EM
角EFM=角FEM=60度
因为角CAB+角CBA+角ACB=180度
因为角CAB=角CBA=80度
所以角ACB=20度
所以角FCA=角FCB-角ACB=60-20=40度
因为AC=BC
所以AC=FB
FC=AC
所以角CFA=角FAC
因为角FCA+角FAC+角CFA=180度
所以角CFA=角FAC=70度
所以角AFB=角CFA-角CFB=70-60=10度
因为角CAE=10度
所以角AFB=角CAE
因为角CBA=80度
角FBC=60度
所以角FBA=角CBA-角FBC=80-60=20度
所以角FBA=角ACB=20度
因为FB=AC(已证）
所以三角形FAB和三角形ACE全等（ASA)
所以AF=AE
所以角AFE=角AEF
因为角FAE=角FAC+角CAE=70+10=80度
角FAE+角AFE+角AEF=180度
所以角AFE=角AEF=50度
因为AG=AG
FM=EM (已证）
AF=AE（已证）
所以三角形FAM和三角形EAM全等（SSS)
所以角AMF=角EAM=1/2角FME=1/2*60=30度
因为角FBA=角FBC-角DBC=60-20=40度
所以角FBD=角FCD=40度
角ACB=角DBC=20度
所以DC=DB
因为FC=FB（已证）
所以三角形FDB和三角形FDC全等（SAS)
所以角DFB=角DFC=1/2角CFB=1/2*60=30度
因为角FDA=角FCA+角DFC=40+30=70度
所以角FDA=角FAD=70度
所以FA=FD
角AFD=180-角FAD-角ADF=40度
所以角DFE=角AFE-角AFD=50-40=10度
所以角DFE=角AFM=10度
因为FE=FM（已证）
所以三角形FAM和三角形FDE全等（SAS)
所以角AMF=角DEF=30度
因为角AEF=角DEA+角DEF=50度（已证）
所以角DEA=50-30=20度
证明过程如下图：
由图可得，可以将EA与BD两线的交点设为F，由三角形角定理可的，内角和为180度，在三角形ABF中，已知∠FAB=70°，∠FBA=60°，那么∠AFB=50°，∠ADB=40°，由对角定律可得∠DFE=∠AFB=50°，∠AFD=180-50=130°，∠ADF=40°，∠CDB=140°，由外角定律，∠AEB=30°，由对称定理可获得∠CDE=30°，∠EDF=180-∠ADB-∠CDE=110°，由三角形内角和定理，求得∠DEA=180°-50°-110°=20°
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