?$???(2) 解：設D(m，m^2-2m-3)，連接OD則0＜m＜3，m^2-2m-3＜0 ???$?
?$???且S_{△AOC}=\frac{3}{2} ???$?
$S_{△DOC}=\frac {3}{2}mS_{△DOB}$
$?$$?=-\frac{3}{2}(m^2-2m-3)?$$?$
?$???∴S_{四邊形ABDC}=S_{△AOC}+S_{△DOC}+S_{△DOB} ???$?
?$???=-\frac{3}{2}m^2+\frac{9}{2}m+6???$?
?$???=-\frac{3}{2}(m-\frac{3}{2})^2+\frac{75}{8}???$?
?$??∴存在點D(\frac{3}{2}，-\frac{15}{4})\ ??$?
?$???使四邊形ABDC的面積最大，且最大值為\frac{75}{8} ???$?
(3) 因為?$???S_{\triangle APC}=S_{\triangle APB}，???$??$???\triangle APC???$?與?$???\triangle APB???$?有相同
?$的底?$?$???AP，???$?$?所以?$?$???B，???$?$??$?$???C???$?$?到直線?$?$???AP???$?$?的距離相等。$?
① 當?$???BC// AP???$?時，
?$???k_{BC}=\frac{0 - (-3)}{3 - 0}=1，???$?設直線?$???AP???$?的方程為?$???y=x + m，???$?
把?$???A(-1,0)???$?代入得?$???0=-1 + m，???$??$???m = 1，???$?即?$???y=x + 1。???$?
聯立?$???\begin{cases}y=x + 1\\y=x^{2}-2x - 3\end{cases}，?$
$????$?則?$???x + 1=x^{2}-2x - 3，???$??$???x^{2}-3x - 4=0，???$??$???(x - 4)(x+1)=0，???$?
解得?$???x_1 = 4，???$??$???x_2=-1???$?（舍去），當?$???x = 4???$?時，?$???y=4 + 1 = 5，???$??$???P(4,5)。???$?
② 當?$???AP???$?過?$???BC???$?中點?$???(\frac{3 + 0}{2},\frac{0+( - 3)}{2})???$?
?$即?$?$???(\frac{3}{2},-\frac{3}{2})???$?$?時，$?
設直線?$???AP???$?的方程為?$???y=kx + n，???$?把?$???A(-1,0)，???$??$???(\frac{3}{2},-\frac{3}{2})???$?代入
?$???\begin{cases}0=-k + n\\-\frac{3}{2}=\frac{3}{2}k + n\end{cases}，???$?
由?$???n = k???$?代入?$???-\frac{3}{2}=\frac{3}{2}k + n???$?得?$???-\frac{3}{2}=\frac{3}{2}k + k，???$??$???-\frac{3}{2}=\frac{5}{2}k，???$??$???k=-\frac{3}{5}，???$??$???n=-\frac{3}{5}，???$??$???y=-\frac{3}{5}x-\frac{3}{5}。???$?
聯立?$???\begin{cases}y=-\frac{3}{5}x-\frac{3}{5}\\y=x^{2}-2x - 3\end{cases}，???$??$???x^{2}-2x - 3=-\frac{3}{5}x-\frac{3}{5}，???$??$???5x^{2}-10x - 15=-3x - 3，???$??$???5x^{2}-7x - 12=0，???$??$???(5x + 12)(x - 1)=0，?$
$????$?解得?$???x_1=\frac{12}{5}，???$??$???x_2 = 1???$?（舍去），當?$???x=\frac{12}{5}???$?時，?$???y=-\frac{3}{5}\times\frac{12}{5}-\frac{3}{5}=-\frac{36 + 15}{25}=-\frac{51}{25}，???$??$???P(\frac{12}{5},-\frac{51}{25})。???$?
綜上，?$???P???$?點坐標為?$???(4,5)???$?或?$???(\frac{12}{5},-\frac{51}{25})。???$?
