Polygon

Construct a //gram ABCM, join MD.
Let ∠ABM = x, so ∠BCM = 180° - x∠MCD = x - (180° - x) = 2x - 180°As CM = AB = CD, so ΔCMD is an isosceles triangle.∠CMD = ∠CDM = [180° - (2x - 180°)] / 2 = 180° - xAs ∠AMC = x (properties of //gram), therefore,∠AMC + ∠CMD = x + (180° - x) = 180°Therefore, AMD is a straight line (adjacent ∠s are supplementary).Therefore BC // AD

2014-03-20 15:38:03 補充：
Thanks, 郭Sir, 是我打錯了第二行，正確是：

Let ∠ABC = x, so ∠BCM = 180° - x

2014-03-20 18:49:00 補充：
過獎了，這只是一般的做法，沒甚麼特別的。

事實上是精彩～

ღ(｡◕‿◠｡)ღ

Let ∠AB[M] = x, so ∠BCM = 180° - x

Would it be : Let ∠AB[C] = x, so ∠BCM = 180° - x ?

2014-03-20 15:47:53 補充：
你的解答很精彩，所以用心細讀，學習，學習。