《2025年名校课堂九年级数学上册北师大版》

1. 如图,在$△ABC$中,$∠ACB= 90^{\circ }$,CD是斜边AB上的高,$AD= 9,BD= 4$,那么$CD= $6,$AC= $$3\sqrt{13}$.

答案： 6 3 $\sqrt{13}$

2. 如图,点$P_{1},P_{2},P_{3},P_{4}$均在坐标轴上,且$P_{1}P_{2}⊥P_{2}P_{3},P_{2}P_{3}⊥P_{3}P_{4}$.若点$P_{1},P_{2}的坐标分别为(0,-1),(-2,0)$,则点$P_{4}$的坐标为$(8,0)$.

答案： $(8,0)$

3. 如图,已知$∠DAB= ∠EAC,∠ADE= ∠ABC$.求证:
(1)$△ADE\backsim △ABC$.
证明:$\because ∠DAB = ∠EAC, \therefore ∠DAE = ∠BAC$. 又 $\because ∠ADE = ∠ABC, \therefore △ADE \backsim △ABC$.
(2)$\frac {AD}{AE}= \frac {BD}{CE}$.
证明:$\because △ADE \backsim △ABC, \therefore \frac{AD}{AE} = \frac{AB}{AC}$. $\because ∠DAB = ∠EAC, \therefore △ADB \backsim △AEC$. $\therefore \frac{AD}{AE} = \frac{BD}{CE}$.

答案： 证明:
(1) $\because ∠DAB = ∠EAC, \therefore ∠DAE = ∠BAC$. 又 $\because ∠ADE = ∠ABC, \therefore △ADE \backsim △ABC$.
(2) $\because △ADE \backsim △ABC, \therefore \frac{AD}{AE} = \frac{AB}{AC}$. $\because ∠DAB = ∠EAC, \therefore △ADB \backsim △AEC$. $\therefore \frac{AD}{AE} = \frac{BD}{CE}$.

4. (2023·常德)如图1,在$Rt△ABC$中,$∠ABC= 90^{\circ },AB= 8,BC= 6$,D是AB上一点,且$AD= 2$,过点D作$DE// BC$交AC于点E,将$△ADE$绕点A顺时针旋转到图2的位置,则图2中$\frac {BD}{CE}$的值为____$\frac{4}{5}$.

答案： $\frac{4}{5}$

5. 如图,点E,F,G分别在正方形ABCD的边AB,BC,AD上,$AF⊥EG$.若$AB= 5,AE= DG= 1$,则$BF= $$\frac{5}{4}$.

答案： $\frac{5}{4}$