《2025年名校题库八年级数学上册北师大版》

9. (石室联中)如图,在等腰直角△ABC中,AB= AC= 8,∠A= 90°,E是BC边上一点,D是AC边的中点,连接ED,过点E作EF⊥ED,满足ED= EF,连接DF,交BC于点M,将△DEM沿DE翻折,得到△DEN,连接NF,交DE于点P,若BE= 2√2,则PF的长度是______。

答案： $5\sqrt{2}$　　

10. (天府新区期末)在△ABC中,∠BAC= 90°,AB= AC,D是平面内一点(不与点A,B,C重合),连接AD,BD,CD,∠BDC= 90°。将△ADC沿直线AD翻折得到△ADG,连接CG。
(1)如图1,点D在∠ABC内部,BD交AC于点E,F是BD上一点,且BF= CD,连接AF。
①求证：△ABF≌△ACD；
②若AD= √2/2,CD= 1,求点G到直线BC的距离。
(2)如图2,点D在∠BAC内部,试判断BD,AD,CG之间的数量关系,并说明理由。

答案：

(1)①证明:$\because ∠BDC = ∠BAC = 90^{\circ },\therefore ∠ABF+∠BAC+∠AEB = ∠ACD+∠BDC+∠CED = 180^{\circ }$.又$\because ∠AEB = ∠CED,\therefore ∠ABF = ∠ACD$.在$△ABF$和$△ACD$中,$\begin{cases} AB = AC \\ ∠ABF = ∠ACD \\ BF = CD \end{cases}$,$\therefore △ABF≌△ACD(SAS)$.
②解:$\because △ABF≌△ACD,\therefore ∠BAF = ∠CAD,AF = AD,\therefore ∠FAD = ∠CAD+∠FAC = ∠BAF+∠FAC = 90^{\circ },\therefore ∠ADF = 45^{\circ },\therefore FD = \sqrt{2}AD = 1,\therefore ∠ADC = ∠ADF+∠BDC = 135^{\circ },\therefore BD = BF + FD = CD + FD = 2,\therefore BC = \sqrt{BD^{2}+CD^{2}}=\sqrt{5}$.由翻折得$∠ADG = ∠ADC = 135^{\circ },DG = DC = 1,\therefore ∠CDG = 360^{\circ }-∠ADC - ∠ADG = 90^{\circ },\therefore CG = \sqrt{2}CD = \sqrt{2},∠BDC+∠GDC = 180^{\circ },\therefore $点 B,D,G 共线,$\therefore BG = BD + DG = 3$.设点 G 到直线 BC 的距离为 h,则$\frac{1}{2}BC\cdot h=\frac{1}{2}BG\cdot CD$,解得$h = \frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{5}$,即点 G 到直线 BC 的距离为$\frac{3\sqrt{5}}{5}$.
(2)解:$\frac{\sqrt{2}}{2}CG+BD=\sqrt{2}AD$.理由如下:
如图,过点 A 作$AH⊥AD$,交 DC 的延长线于点 H,$\therefore ∠DAH = ∠DAC+∠CAH = 90^{\circ }$.

$\because ∠BAC = ∠BAD+∠DAC = 90^{\circ },\therefore ∠BAD = ∠CAH$.
$\because ∠BDC = 90^{\circ },\therefore ∠ABD+∠ACD = 360^{\circ }-∠BDC - ∠BAC = 180^{\circ }$.
$\because ∠ACH+∠ACD = 180^{\circ },\therefore ∠ABD = ∠ACH$.
在$△ABD$和$△ACH$中,$\begin{cases} ∠BAD = ∠CAH \\ AB = AC \\ ∠ABD = ∠ACH \end{cases}$,$\therefore △ABD≌△ACH(ASA),\therefore BD = CH,AD = AH,\therefore ∠ADH = 45^{\circ },\therefore DH = \sqrt{2}AD$.
由翻折得$∠ADG = ∠ADC = 45^{\circ },DG = DC,\therefore ∠CDG = 90^{\circ },\therefore CG = \sqrt{2}CD$.
$\because CD + CH = DH,\therefore \frac{\sqrt{2}}{2}CG+BD=\sqrt{2}AD$.