18.(6分)求式中x的值:$(2x-1)^{2}= 36$.
答案：解：$2x - 1 = \pm 6$，
$2x - 1 = 6$或$2x - 1 = -6$，
解得$x = \frac{7}{2}$或$x = -\frac{5}{2}$。
$2x - 1 = 6$或$2x - 1 = -6$，
解得$x = \frac{7}{2}$或$x = -\frac{5}{2}$。
19.(6分)直线l如图所示,求直线l的函数表达式.
答案：解：设直线$l$的函数表达式为$y = kx + b$。
$\because$直线$l$经过点$A(-3,-1)$，$B(1,3)$，
$\therefore \begin{cases}-3k + b = -1\\k + b = 3\end{cases}$，解得$\begin{cases}k = 1\\b = 2\end{cases}$
$\therefore$直线$l$的函数表达式为$y = x + 2$。
$\because$直线$l$经过点$A(-3,-1)$，$B(1,3)$，
$\therefore \begin{cases}-3k + b = -1\\k + b = 3\end{cases}$，解得$\begin{cases}k = 1\\b = 2\end{cases}$
$\therefore$直线$l$的函数表达式为$y = x + 2$。
20.(8分)如图,在四边形ABCD中,AC,BD相交于点O,$∠ABC= ∠DCB,∠1= ∠2$.求证:$OB= OC$.
答案：证明：在$\triangle ABC$和$\triangle DCB$中，
$\begin{cases}\angle 1 = \angle 2\\\angle ABC = \angle DCB\\BC = BC\end{cases}$，$\therefore \triangle ABC \cong \triangle DCB(AAS)$，
$\therefore \angle OBC = \angle OCB$，$\therefore OB = OC$。
$\begin{cases}\angle 1 = \angle 2\\\angle ABC = \angle DCB\\BC = BC\end{cases}$，$\therefore \triangle ABC \cong \triangle DCB(AAS)$，
$\therefore \angle OBC = \angle OCB$，$\therefore OB = OC$。
21.(8分)如图,AD是$\triangle ABC$的中线,$DE⊥AC$于点E,DF是$\triangle ABD$的中线,且$CE= 2,DE= 4,AE= 8$.
(1)求证:$∠ADC= 90^{\circ }$;
(2)求DF的长.
(1)求证:$∠ADC= 90^{\circ }$;
(2)求DF的长.
答案：(1) 证明：$\because DE \perp AC$于点$E$，$\therefore \angle AED = \angle CED = 90^{\circ}$，
在$Rt\triangle ADE$中，$\angle AED = 90^{\circ}$，
$\therefore AD^{2} = AE^{2} + DE^{2} = 8^{2} + 4^{2} = 80$，同理，$CD^{2} = 20$，
$\therefore AD^{2} + CD^{2} = 100$，$\because AC = AE + CE = 8 + 2 = 10$，
$\therefore AC^{2} = 100$，$\therefore AD^{2} + CD^{2} = AC^{2}$，
$\therefore \triangle ADC$是直角三角形，$\therefore \angle ADC = 90^{\circ}$。
(2) 解：$\because AD$是$\triangle ABC$的中线，$\angle ADC = 90^{\circ}$，$\therefore AD$垂直平分$BC$，$\therefore AB = AC = 10$，在$Rt\triangle ADB$中，$\angle ADB = 90^{\circ}$，$\because DF$是$\triangle ABD$的中线，$\therefore DF = \frac{1}{2}AB = 5$。
在$Rt\triangle ADE$中，$\angle AED = 90^{\circ}$，
$\therefore AD^{2} = AE^{2} + DE^{2} = 8^{2} + 4^{2} = 80$，同理，$CD^{2} = 20$，
$\therefore AD^{2} + CD^{2} = 100$，$\because AC = AE + CE = 8 + 2 = 10$，
$\therefore AC^{2} = 100$，$\therefore AD^{2} + CD^{2} = AC^{2}$，
$\therefore \triangle ADC$是直角三角形，$\therefore \angle ADC = 90^{\circ}$。
(2) 解：$\because AD$是$\triangle ABC$的中线，$\angle ADC = 90^{\circ}$，$\therefore AD$垂直平分$BC$，$\therefore AB = AC = 10$，在$Rt\triangle ADB$中，$\angle ADB = 90^{\circ}$，$\because DF$是$\triangle ABD$的中线，$\therefore DF = \frac{1}{2}AB = 5$。