27. (12分)图形变换可以帮助我们认识图形.
(1)把图1中等腰三角形纸片沿着顶角平分线折叠得到图2,由△ABD与△ACD重合,可知∠B=∠

C

,BD=

CD

;
(2)如图3,将△ABC沿边AB的垂直平分线翻折得到△BAD,点A对应点B,点C对应点D,再将△BAD绕点B逆时针旋转得到△BA'D',当点A'恰好落在AC的延长线上时,判断BD'与AC的数量关系和位置关系,并说明理由;
(3)如图4,△ABC中,∠BAC=90°,∠B=α°,点D在AB上,过点D作DE//BC交AC于点E,将所截△ADE沿过点A的某射线AF翻折得到△AD'E'.直接写出当△AD'E'的某一边与BC平行时∠BAF的大小.(只写出∠BAF为锐角时的大小即可,结果用含α的代数式表示)

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27.【点拨】本题考查平行线的性质,轴对称的性质,旋转的性质等知识,解决问题的关键是分类讨论.
【解析】(1)$\because$ 点B和点C重合,$\therefore ∠ B = ∠ C$,$BD=CD$. 故答案为$C$,$CD$.
(2)$BD' = AC$,$BD'// AC$. 理由如下:
$\because △ ABC$沿边$AB$的垂直平分线翻折得到$△ BAD$,
$\therefore BD = AC$,$∠ ABD = ∠ BAC$.
$\because △ BAD$绕点$B$逆时针旋转得到$△ BA'D'$,
$\therefore BD' = BD$,$∠ A'BD' = ∠ ABD$,$A'B = AB$,
$\therefore BD' = AC$,$∠ BA'C = ∠ BAC$,
$\therefore ∠ A'BD' = ∠ BA'C$,
$\therefore BD'// AC$.
(3)如图1,当$D'E'// BC$,$DE$和$D'E'$在同一直线上时.
$\because DE// BC$,$\therefore$ 点$D'$在$DE$上,
$\therefore AF⊥ DE$.
$\because DE// BC$,$\therefore AF⊥ BC$,
$\therefore ∠ BAF = 90° - ∠ B = 90° - α°$;
如图2,当$D'E'// BC// DE$时,
$∠ BAF = ∠ B = α°$;
如图3,当$AE'// BC$,点$E'$在$AB$的左侧时,
$∠ BAE' = ∠ B = α°$.
$\because ∠ D'AE' = ∠ DAE = 90°$,
$\therefore ∠ BAD' = 90° - ∠ BAE' = 90° - α°$,
$\therefore ∠ BAF = ∠ D'AF = \frac{1}{2}∠ BAD' = 45° - \frac{1}{2}α°$;
如图4,当$AD'// BC$,点$D'$在$AB$左侧时,
$∠ BAF = ∠ D'AF = \frac{1}{2}∠ BAD' = \frac{1}{2}∠ B = \frac{1}{2}α°$;
如图5,当$AD'// BC$,点$D'$在$AC$的右侧时,
$∠ DAD' = 180° - ∠ B = 180° - α°$,
$\therefore ∠ BAF = ∠ D'AF = \frac{1}{2}∠ DAD' = 90° - \frac{1}{2}α°$;
如图6,当$AE'// BC$,点$E'$在$AC$的右侧时,
$∠ CAE' = ∠ ACB = 90° - α°$,
$\therefore ∠ CAG = \frac{1}{2}∠ CAE' = 45° - \frac{1}{2}α°$,
$\therefore ∠ BAF = 90° - ∠ CAG = 45° + \frac{1}{2}α°$.
综上所述,$∠ BAF = 90° - α°$或$α°$或$\frac{1}{2}α°$或$45° - \frac{1}{2}α°$或$90° - \frac{1}{2}α°$或$45° + \frac{1}{2}α°$.