《2025年名师帮同步学案九年级数学全一册人教版》

1.如图,在$\triangle ABC$中,$DE// BC,AD= 2,BD= 3,$$AC= 10$,则AE的长为____4.

答案： 4

2.如图,在$\triangle ABC$的AB边上取点D,作DE交AC于点E,且$∠AED= ∠B$.若$AB= 4,AD= 2,$$AC= 3$,则EC的长为____$\frac {1}{3}$.

答案： $\frac {1}{3}$

3.如图,在$\triangle ABC$中,$∠ABD= ∠C$.若$AB= 4,AD=2$,则CD的长为____6.

答案： 6

4.如图,AB为$\odot O$的直径,CD为$\triangle ABC$的高,$BD= 4,CD= 6$,则$\odot O$的直径长为____13.

答案： 13

5.如图,点E在平行四边形ABCD的边DC上,若$DE:EC= 2:3$,则$AF:FC= $$5:3$.

答案： $5:3$

6.如图,在$\odot O$中,弦AC,BD相交于点E.若$AE= 4,CE= 3,BD= 8$,则BE的长为____2 或 6.

答案： 2 或 6

7.如图,已知$A(0,4),B(4,1),BC⊥x$轴于点C,点P为线段OC上一点,且$PA⊥PB$,则点P的坐标为(B)

A.$(1,0)$
B.$(2,0)$
C.$(\frac {3}{2},0)$
D.$(\frac {5}{2},0)$

答案： B

8.如图,已知$∠B= ∠ACE= ∠D$,有下列四个结论:①$∠BAC= ∠DCE$;②$∠ACB= ∠CED;$③$\triangle ABC\backsim \triangle CDE$;④$\frac {AC}{BC}= \frac {CE}{CD}$.其中正确的有(C)

A.①②④
B.①③④
C.①②③
D.①②③④

答案： C

9.如图,在$\triangle ABC$中,$∠BAC= 90^{\circ },AB= AC$,点D是BC边上的一个动点(不与点B,C重合),$∠ADE= 45^{\circ }$.
(1)求证:$\triangle ABD\backsim \triangle DCE$;
(2)若$AB= 4,BD= \sqrt {2}$,求EC的长.$\frac {3}{2}$

答案：
(1)证明:$\because ∠BAC=90^{\circ },AB=AC,$
$\therefore ∠B=∠C=45^{\circ }.$
又$\because ∠ADB=∠DAC+∠C=∠DAC+45^{\circ },$
$∠DEC=∠DAC+∠ADE=∠DAC+45^{\circ },$
$\therefore ∠ADB=∠DEC,$
$\therefore △ABD\backsim △DCE;$
(2)解:$\because ∠BAC=90^{\circ },AB=AC,AB=4,$
$\therefore BC=\sqrt {AB^{2}+AC^{2}}=4\sqrt {2}.$
$\because BD=\sqrt {2},\therefore DC=3\sqrt {2}.$
$\because △ABD\backsim △DCE,$
$\therefore \frac {EC}{BD}=\frac {DC}{AB},$
$\therefore EC=\frac {3}{2}.$

10.如图，已知∠DAB=∠EAC，∠ADE=∠ABC.求证:AD· CE=AE· BD.
证明:∵∠DAB=∠EAC,∴∠DAB+∠BAE=∠EAC+∠BAE,即∠DAE=∠BAC.∵∠ADE=∠ABC,∴△ADE∽△ABC,∴ADAB=AEAC.又∵∠DAB=∠EAC,∴△ABD∽△ACE,∴ADAE=BDCE,∴AD· CE=AE· BD.

答案： 10.证明:$\because ∠DAB=∠EAC,$ $\therefore ∠DAB+∠BAE=∠EAC+∠BAE,$ 即$∠DAE=∠BAC.$ $\because ∠ADE=∠ABC,$ $\therefore △ADE\backsim △ABC,\therefore \frac {AD}{AB}=\frac {AE}{AC}.$ 又$\because ∠DAB=∠EAC,$ $\therefore △ABD\backsim △ACE,$ $\therefore \frac {AD}{AE}=\frac {BD}{CE},\therefore AD\cdot CE=AE\cdot BD.$