《2025年名校题库八年级数学上册北师大版》

9. (武侯区期末)如图,在平面直角坐标系xOy中,直线$y= x+2+\sqrt {2}$交x轴于点A,过该直线上一点B作$BC⊥y$轴于点C,且$OC= 2\sqrt {2}$.
(1)求点B的坐标及线段AB的长.
(2)取OC的中点D,作直线BD交x轴于点E,连接AD.
①求证:AD是$∠BAE$的平分线.
②若M,N分别是线段AO,AD上的动点,连接MN,ON,试问:$MN+ON$是否存在最小值?若存在,求出该最小值;若不存在,请说明理由.

答案：

(1)解:$\because OC = 2\sqrt{2}$,$\therefore y_{B}=2\sqrt{2}$.在$y = x + 2+\sqrt{2}$中,令$y = 2\sqrt{2}$,得$x=\sqrt{2}-2$,$\therefore B(\sqrt{2}-2,2\sqrt{2})$.在$y = x + 2+\sqrt{2}$中,令$y = 0$,得$x=-2-\sqrt{2}$,$\therefore A(-2-\sqrt{2},0)$,$\therefore AB=\sqrt{(\sqrt{2}-2 + 2+\sqrt{2})^{2}+(2\sqrt{2}-0)^{2}}=4$.
(2)①证明:$\because D$是$OC$的中点,$\therefore D(0,\sqrt{2})$,$CD = OD$.设直线$BD$的解析式为$y = kx+\sqrt{2}$,把点$B(\sqrt{2}-2,2\sqrt{2})$代入,得$2\sqrt{2}=(\sqrt{2}-2)k+\sqrt{2}$,解得$k=-(1+\sqrt{2})$,$\therefore$直线$BD$的解析式为$y=-(1+\sqrt{2})x+\sqrt{2}$.在$y=-(1+\sqrt{2})x+\sqrt{2}$中,令$y = 0$,得$x = 2-\sqrt{2}$,$\therefore E(2-\sqrt{2},0)$,$\therefore AE = 2-\sqrt{2}-(-2-\sqrt{2})=4$.由
(1)知$AB = 4$,$\therefore AB = AE$,即$\triangle ABE$是等腰三角形.$\because CD = OD$,$\angle BDC=\angle EDO$,$\angle BCD=\angle EOD = 90^{\circ}$,$\therefore\triangle BCD\cong\triangle EOD(ASA)$,$\therefore BD = ED$,$\therefore AD$是$\angle BAE$的平分线. ②解:$MN + ON$存在最小值.作点$O$关于$AD$的对称点$H$,连接$DH$,如图所示.由①知$AD$是$\angle BAE$的平分线,$\therefore$点$H$在线段$AB$上.$\because$点$N$在$AD$上,$\therefore ON = HN$,$\therefore MN + ON=MN + HN$,$\therefore$当$H$,$N$,$M$三点共线,且$HM\perp OA$时,$MN + HN$的值最小,即$MN + ON$的值最小,$HM$的长即是$MN + ON$的最小值.由对称性可得,$AH = OA = 2+\sqrt{2}$.$\because$直线$y = x + 2+\sqrt{2}$与$x$轴的夹角为$45^{\circ}$,即$\angle HAM = 45^{\circ}$,$\therefore\triangle AHM$是等腰直角三角形,$\therefore HM=\sqrt{2}+1$,$\therefore MN + ON$的最小值是$\sqrt{2}+1$.