求解一道数学几何题

不用像楼上mouvca2那样线段乘来乘去比来比去，只用△等腰等底角和△内角和为180度这样的性质来进行角度加减计算就好。关键是分割出适当的等腰△。以下将内角为80°，20°，80°的三角形称为8-2-8△，其它整10倍角的三角形亦然。

【解】△ABC为8-2-8△，作顶角C的平分线CF交BD于F，将BD沿CF反射为AG，连结DG。那么

①△ABF为等边△，②△DFG为等边△，③△ACG为2-14-2△

由CF、AE同是△ACG的底角平分线，所以连结EF得

④四边形ACEF为等腰梯形，⑤△EFG为2-14-2△，⑥△HEF为1-16-1△

由②和⑤得

⑦△DEG为5-8-5△

最后得，x=50°-10°-20°=20°

你是几年级？看可以用那些知识解

计算器没有还有三角变换忘光了，你自己算算看吧答案对不对，这方法应该是对的 

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1.20 ，几何方法求没求出来，
猜想20，可用同一法证明，
2.用三角函数：BA=1/2*BC*sin10° ①
计算：CD/DA=三角形CDB的面积/三角形ADB的面积=（1/2*BD*BC*sin20°）/（1/2*BD*BA*sin600°）=1/(4sin20°sin60°) ②
在三角形ABE中易知AB/AE=sin30°/sin80°，即BA=AE*1/(2cos40°) ③

法一: 几何方法

过点D作DH//AB，交BC于点H。连接DB, HA, DB交HA于点M.
易证:
(1). AM = MB = BA, DM = MH = HD,
(2). ∠EAH = 10度 =∠CAE. CD = BD = CH = AH

(1), (2) ==> EH/EC = AH/AC ==> CH/CE = AH/CE = (AH + AC)/AC = (AH + BC)/BC
==> AH*BC = CE*(AH +BC) ==> AH*BE = CE*BC, AH/BC = CE/BE
==> MD/MB = DH/AB = CH/BC = AH/BC = CE/BE ==> ME // DC
==> MH = EH = DH ==> ∠HDE = ∠HED = 50°
==> x = ∠DEA = ∠HED - ∠HEA = 50° - 30° = 20°

Answer: x = 20°

法2: A lazy way.

∠AEB = 30°,
Let AB = 1.
By Sine Law,
AD/sin60° = AB/sin40° = 1/sin40°, AD = sin60°/sin40°.
AE/sin80° = AB/sin30°, AE = sin80°/sin30° = 2sin80°
By Cosine Law,
DE^2 = AD^2 + AE^2 - 2AD*AE*∠DAE
= (sin60°/sin40°)^2 + 4(sin80°)^2 - 4sin60°sin80°cos10°/sin40°
= 0.467911113...
DE = 0.684040286...
By Sine Law,
sinx / AD = sin10°/DE
sinx = ADsin10°/DE = sin60°sin10°/(DE*sin40°) = 0.342020143...
x = arcsin(0.342020143...)
= 20°.

设角BDE为y,AE'BD交于O点。.角C 为20度，角BDA为40度。角CDB为140度。角DOE为130度则 :x+y=50，x+10=180-40+y.解得；X=90