MATH F.2 20點..要快..

BE=EC(given) and EC=ED(given)
∠EBC=62(sides opp. equal ∠s) and
∠D=∠DCE(sides opp. equal ∠s)
AE//CD(given)


∠BCE+62+62=180(∠sum of △)
∠BCE=56

2∠D+48=180
∠D=66


(x+56+48)+66=180(int.∠s,AE///CD)
x=10

2009-06-06 12:02:03 補充：
以上全部unit(degree,°)略去。

2009-06-06 12:02:51 補充：
F.2不可以這樣跳步....
會被扣pp

(I am talking about #1 answer)

Triangle CDE is an isos. triangle. Therefore angle CDE = (180 - 48)/2 = 66.
Triangle BCE is also an isos. triangle, therefore angle CEB = 180 - 62 - 62 = 56.
Since AE//CD
x + angle CEB + angle CED + angle CDE = 180 ( suppl. angles AE//CD)
x + 56 + 48 + 66 = 180
x = 10 degree.