22.(10分)如图,直线$l_{1}:y= x+1$与x轴、y轴分别交于点A,B,另一直线$l_{2}:y= -\frac {3}{4}x+b$与x轴、y轴分别交于点C,D,连接AD,直线$l_{1}与直线l_{2}交于点E(2,m)$,在x轴上有一点$P(a,0)$(其中$a＞2$),过点P作x轴的垂线,分别与直线$l_{1},l_{2}$交于点M,N.
(1)求b的值及$\triangle ADE$的面积;
(2)若$MN= BD$,求a的值.
(1)求b的值及$\triangle ADE$的面积;
(2)若$MN= BD$,求a的值.
答案：解：(1)$\because$直线$l_{1}:y = x + 1$经过点$E(2,m)$，
$\therefore m = 2 + 1 = 3$，$\therefore E(2,3)$，
把点$E$的坐标代入$y = -\frac{3}{4}x + b$得，$3 = -\frac{3}{4}×2 + b$，
解得$b = \frac{9}{2}$，$\therefore$直线$l_{2}:y = -\frac{3}{4}x + \frac{9}{2}$。
$\because$直线$l_{1}:y = x + 1$与$x$轴、$y$轴分别交于点$A$，$B$，直线$l_{2}:y = -\frac{3}{4}x + \frac{9}{2}$与$x$轴、$y$轴分别交于点$C$，$D$，
$\therefore A(-1,0)$，$B(0,1)$，$D(0,\frac{9}{2})$，$\therefore BD = \frac{7}{2}$，$\therefore \triangle ADE$的面积$= S_{\triangle ABD} + S_{\triangle BDE} = \frac{1}{2}×\frac{7}{2}×(2 + 1) = \frac{21}{4}$。
(2)$\because MN // y$轴，$\therefore \angle DBE = \angle EMN$，
在$\triangle DBE$和$\triangle NME$中，
$\begin{cases}\angle DBE = \angle EMN\\\angle BED = \angle MEN\\BD = MN\end{cases}$，$\therefore \triangle DBE \cong \triangle NME(AAS)$，
$\therefore BE = ME$，
$\because$点$E(2,m)$，$\therefore$点$M$的横坐标为$4$，$\therefore a$的值为$4$。
$\therefore m = 2 + 1 = 3$，$\therefore E(2,3)$，
把点$E$的坐标代入$y = -\frac{3}{4}x + b$得，$3 = -\frac{3}{4}×2 + b$，
解得$b = \frac{9}{2}$，$\therefore$直线$l_{2}:y = -\frac{3}{4}x + \frac{9}{2}$。
$\because$直线$l_{1}:y = x + 1$与$x$轴、$y$轴分别交于点$A$，$B$，直线$l_{2}:y = -\frac{3}{4}x + \frac{9}{2}$与$x$轴、$y$轴分别交于点$C$，$D$，
$\therefore A(-1,0)$，$B(0,1)$，$D(0,\frac{9}{2})$，$\therefore BD = \frac{7}{2}$，$\therefore \triangle ADE$的面积$= S_{\triangle ABD} + S_{\triangle BDE} = \frac{1}{2}×\frac{7}{2}×(2 + 1) = \frac{21}{4}$。
(2)$\because MN // y$轴，$\therefore \angle DBE = \angle EMN$，
在$\triangle DBE$和$\triangle NME$中，
$\begin{cases}\angle DBE = \angle EMN\\\angle BED = \angle MEN\\BD = MN\end{cases}$，$\therefore \triangle DBE \cong \triangle NME(AAS)$，
$\therefore BE = ME$，
$\because$点$E(2,m)$，$\therefore$点$M$的横坐标为$4$，$\therefore a$的值为$4$。
23.(10分)如图,$\triangle ABC$中,$AD⊥BC$,垂足为D,$BD= 1,AD= 2,CD= 4$.
(1)求证:$∠BAC= 90^{\circ }$;
(2)点P为BC上一点,连接AP,若$\triangle ABP$为等腰三角形,求BP的长.
(1)求证:$∠BAC= 90^{\circ }$;
(2)点P为BC上一点,连接AP,若$\triangle ABP$为等腰三角形,求BP的长.
答案：(1) 证明：$\because AD \perp BC$，$AD = 2$，$BD = 1$，
$\therefore AB^{2} = AD^{2} + BD^{2} = 5$，
又$\because AD \perp BC$，$CD = 4$，$AD = 2$，$\therefore AC^{2} = CD^{2} + AD^{2} = 20$，
$\because BC = CD + BD = 5$，$\therefore BC^{2} = 25$，
$\therefore AC^{2} + AB^{2} = 25 = BC^{2}$，
$\therefore \angle BAC = 90^{\circ}$。
(2) 解：分三种情况：
①当$BP = AB$时，$\because AD \perp BC$，$\therefore AB = \sqrt{BD^{2} + AD^{2}} = \sqrt{5}$，$\therefore BP = AB = \sqrt{5}$；
②当$BP = AP$时，$P$是$BC$的中点，$\therefore BP = \frac{1}{2}BC = 2.5$；
③当$AP = AB$时，$BP = 2BD = 2$。
综上所述，$BP$的长为$\sqrt{5}$或$2$或$2.5$。
$\therefore AB^{2} = AD^{2} + BD^{2} = 5$，
又$\because AD \perp BC$，$CD = 4$，$AD = 2$，$\therefore AC^{2} = CD^{2} + AD^{2} = 20$，
$\because BC = CD + BD = 5$，$\therefore BC^{2} = 25$，
$\therefore AC^{2} + AB^{2} = 25 = BC^{2}$，
$\therefore \angle BAC = 90^{\circ}$。
(2) 解：分三种情况：
①当$BP = AB$时，$\because AD \perp BC$，$\therefore AB = \sqrt{BD^{2} + AD^{2}} = \sqrt{5}$，$\therefore BP = AB = \sqrt{5}$；
②当$BP = AP$时，$P$是$BC$的中点，$\therefore BP = \frac{1}{2}BC = 2.5$；
③当$AP = AB$时，$BP = 2BD = 2$。
综上所述，$BP$的长为$\sqrt{5}$或$2$或$2.5$。