8.如图,在四边形ABCD中,AB//CD,AC平分∠BAD,CE//AD交AB于E.
(1)求证:四边形AECD是菱形；
(2)若点E是AB的中点,试判断△ABC的形状,并说明理由.

8.(1)证明:$\because AB// CD$,
即$AE// CD$,
又$\because CE// AD$,
$\therefore$ 四边形AECD是平行四边形.
$\because AC$平分$∠ BAD$,
$\therefore ∠ CAE=∠ CAD$.
又$\because AD// CE$,
$\therefore ∠ ACE=∠ CAD$.
$\therefore ∠ ACE=∠ CAE$.
$\therefore AE=CE$.
$\therefore$ 四边形AECD是菱形.
(2)解法一:$\because E$是$AB$的中点,
$\therefore AE=BE$.
又$\because AE=CE$,
$\therefore BE=CE$.
$\therefore ∠ B=∠ BCE$.
$\because ∠ B+∠ BCA+∠ BAC=180°$,
$\therefore 2∠ BCE+2∠ ACE=180°$.
$\therefore ∠ BCE+∠ ACE=90°$.
即$∠ ACB=90°$.
$\therefore △ ABC$是直角三角形.
解法二:连接$DE$,则$DE⊥ AC$,且$DE$平分$AC$,
设$DE$交$AC$于$F$.
$\because E$是$AB$的中点,
$\therefore EF// BC$.
$\therefore BC⊥ AC$,
$\therefore △ ABC$是直角三角形.

9.如图,已知直线$y=2x+3$与直线$y=-2x-1$.
(1)求两直线与y轴的交点A,B的坐标;
(2)求两直线交点C的坐标;
(3)求$△ ABC$的面积.

9.(1)$A(0,3),B(0,-1)$.
(2)由$\begin{cases} y=2x+3,\\ y=-2x-1 \end{cases}$
解得$\begin{cases} x=-1,\\ y=1. \end{cases}$
$\therefore C(-1,1)$.
(3)2.