數學高手請進-----數學大難題 20分

AB同CD相交個點叫F
In △ABC,
AB = AC (given)
∠BAC+∠ABC+∠ACB=180° (∠sum of △)
20°+∠ABC+∠ACB=180°
∠ABC+∠ACB=160°
∠ABC=∠ACB=80° (base ∠s, isos. △)

In △BFC,
∠DBC=∠FBC+∠DBF
80°=∠FBC+20°
∠FBC=60°

In △BCE,
∠EBC+∠ECB+∠BEC=180° (∠sum of △)
60°+80°+∠BEC=180°
∠BEC=40°

In △ACD,
∠DAC+∠DCA+∠ADC=180° (∠sum of △)
20°+30°+∠ADC=180°
∠ADC=130°

Let AB=AC=x
BC²=AB²+AC²-2AB∙AC cos∠BAC
x²+x²-2x∙x cos20°
2x²-2x²cos20°
2x²(1-cos20°)
BC=√[2x²(1-cos20°)]
≈0.3473x
BC/sin∠BEC=CE/sin∠CBE
CE=BC sin∠CBE/sin∠BEC
≈0.3473x sin60°/sin40°
≈0.4679x
AC/sin∠ADC=CD/sin∠DAC
CD=AC sin∠DAC /sin∠ADC
=x sin20°/sin130°
≈0.4465x
DE²=CD²+CE²-2CD∙CE cos ∠DCE
=0.4465x²+0.4679x²-2*0.4465x*0.4679x cos 30°
≈0.0564x²
DE ≈0.2376x
cos∠DEC=(DE²+CE²-CD²)/2CE∙DE
≈[0.0564x²+(0.4679x)²-(0.4465x)²]/2(0.4679x∙0.2376x)
≈(0.0564x²+0.2189x²-0.1993x²)/2*0.1112x²
≈0.076x²/0.2223x²
≈0.342
∠DEC=70°

設BC有一條直線,中間就會變成一隻對頂角,之後就會劃出了4個三角,由對頂角中,須時針得出上下,二角是70,左,右三角是110,再加以連算得出DEC< 是40度

2007-12-31 00:09:57 補充：
再加上去總ans是80

LET 多樣先,AB同CD相交個點叫O

角ADC+角A+角C=180(3角形內角和)
角ADC=180-20-30=130
角AEB+角A+角B=180(3角形內角和)
角AEB=180-20-20=140
角A+角AED+角ADE=180(3角形內角和)
(角AED+角ADE)=180-20=160
角ADC+角AEB=(角AED+角ADE)+角DEB+角EDC
角DEB+角EDC=角ADC+角AEB-(角AED+角ADE)=130+140-(160)=110
角DOE+角DEB+角EDC=180(3角形內角和)
角DOE=180-110=70

計到係咁多..DATA唔夠掛
角DOE+角EOC=180(直線上的鄰角)
角EOC=180-70=110
角CEO+角EOC+角C=180(3角形內角和)
角CEO=180-110-30=40

a162 b135 c121 d142 e140
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...好唔清楚呀你d問題