Plane Geometry

(1) See explanation from ㊟ 受襲貓 ㊣㊙㊡.
(2) Cannot be found because of not enough information.
(3) Let angle BCD = x
angle CBD = x ( base angle of isos. triangle)
angle BDE = 2x ( ext. angle of triangle)
angle BED = 2x ( base angle of isos. triangle)
angle BAD = angle BEA = 60 degree ( angle of equilateral triangle)
angle BCD + angle BED + angle BEA + angle BAD = 180 degree ( angle sum of triangle)
x + 2x + 60 + 60 = 180
3x = 60
x = angle BCD = 20 degree.
(4) Angle BAC = 30 degree ( alt. angle AB//CD).
Angle BCA = angle BAC = 30 degree ( base angle of isos. triangle)
Angle BAC + angle BCA + angle ABE + angle CBD = 180 degree ( angle sum of triangle)
so 30 + 30 + 60 + angle CBD = 180
angle CBD = 60 degree.

2013-09-29 06:10:37 補充：
For Q2, is it BE = CD, not BC = CD ?

2013-09-30 08:21:11 補充：
Q3. Sorry, should be angle BAE, not BAD.
Q2.X and Y are 2 points on BC such that EX and DY are perpendicular to BC.Triangle BEX is congruent to triangle DYC (RHS). So angle EBX = angle DCY. Angle EBX = 90 - (115 - 90) = 180 - 115 = 65 degree = angle BCD. Angle A = angle C = 65 degree.

2013-09-30 08:21:52 補充：
Angle ABE = 115 - 65 = 50 degree ( ext. angle of triangle).

1)
ASA 的邊必須於兩角之間
AAS 的邊必須不是兩角之間的邊

但由於 ∠ sum of △，其實如果你多計算出那隻餘下的角，那麼你喜歡用ASA或AAS均可，但你的算式仍需配合是ASA或AAS的情況，視乎你的步驟如何寫。