(2)
:
畫出△CD′D″(A),
①平行,
理由:∠DCE=∠DCE′=∠D′CA=∠α,
∴∠BAC=∠D′CA=∠α,
∴AB∥CD′.
②∵四邊形ABCD′是等腰梯形,
∴∠ABC=∠D′AB=2∠BAC=2∠α,
∵AB=AC,
∴∠ABC=∠ACB=2∠α,
在△ABC中,∠A+∠ACB+∠ABC=180°,
解之得∠α=36°.點(diǎn)評(píng):本題綜合考查了軸對(duì)稱圖形及旋轉(zhuǎn)變換作圖及平行線的判定和三角形的內(nèi)角和,學(xué)生對(duì)所學(xué)過的知識(shí)要統(tǒng)一起來.
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