$解：? (1)?二次函數(shù)的對稱軸為?x=-\frac {2a} = 1?$
$當(dāng)?x=1?時，?y=a-\frac {1}{2}?$
∵頂點在一次函數(shù)的圖像上
$∴頂點縱坐標(biāo)為?y=-2×1=-2?$
$∴?a-\frac {1}{2}=-2?$
$∴?a=-\frac {3}{2}?$
$?(2)?二次函數(shù)表達式為?y=\frac {1}{2}x2-x-\frac {3}{2}?$
$令?y=0，??\frac {1}{2}x2-x-\frac {3}{2}=0?$
$解得?{x}_1=-1，??{x}_2=3?$
$∴?A(-1，??0)，??B(3，??0)?$
$?(3)?∵四邊形?ACBD?是平行四邊形$
$∴點?C、??D?關(guān)于對角線交點?(1，??0)?對稱$
$又∵點?D'?是點?D?關(guān)于?x?軸的對稱點$
$∴點?C、??D'?關(guān)于拋物線的對稱軸(直線?x=1 )?對稱$
$∴點?D'?在二次函數(shù)圖像上$
