《2025年零障碍导教导学案九年级数学全一册人教版》

7. 如图,AD 与 BC 交于点 E,$AB// CD,AB= 6,CD= 8,BE= 3$,求 BC 的长.
解: ∵AB//CD,
∴∠A = ∠D, ∠B = ∠C.
∴△ABE∽△DCE.
∴$\frac{AB}{DC}$ = $\frac{BE}{CE}$, 即CE = 4.
∴BC = BE + EC = 3 + 4 = 7.

答案： 解:
∵AB//CD,
∴∠A = ∠D, ∠B = ∠C.
∴△ABE∽△DCE.
∴$\frac{AB}{DC}$ = $\frac{BE}{CE}$, 即CE = 4.
∴BC = BE + EC = 3 + 4 = 7.

8. 如图,下列不能判定$△AED与△ABC$相似的是(D)

A. $∠1= ∠C$
B. $∠2= ∠B$
C. $\frac {AE}{AC}= \frac {AD}{AB}$
D. $\frac {AE}{AC}= \frac {ED}{BC}$

答案： D

9. 如图,AD,BC 交于点 O,P 为 AB,CD 延长线的交点,且$PA\cdot PB= PC\cdot PD.$
求证:$△PAD\backsim △PCB.$

证明: ∵PA·PB = PC·PD,
∴$\frac{PA}{PC}$ = $\frac{PD}{PB}$.
∵∠P = ∠P,
∴△PAD∽△PCB.

答案： 证明:
∵PA·PB = PC·PD,
∴$\frac{PA}{PC}$ = $\frac{PD}{PB}$.
∵∠P = ∠P,
∴△PAD∽△PCB.

10. 如图,$△ABC内接于\odot O$,AD 是边 BC 上的高,AE 是$\odot O$的直径.
求证:$AB\cdot AC= AD\cdot AE.$

答案：
证明: 如图, 连接BE,

∵AE为⊙O的直径,
∴∠ABE = 90°.
∵AD⊥BC,
∴∠ADC = 90°.
∴∠ABE = ∠ADC = 90°.
∵$\overparen{AB}$ = $\overparen{AB}$,
∴∠E = ∠C.
∴△ABE∽△ADC.
∴$\frac{AB}{AD}$ = $\frac{AE}{AC}$.
∴AB·AC = AD·AE.

11. 如图,四边形 ABCD,DCFE,EFGH 都是相同的正方形.
(1)求证:$△ACF\backsim △GCA;$
(2)求$∠1+∠2$的度数.

(1) 证明: 设正方形的边长为a, 则
AC = $\sqrt{a^2 + a^2}$ = $\sqrt{2}a$.
∵$\frac{CF}{CA}$ = $\frac{a}{\sqrt{2}a}$ = $\frac{1}{\sqrt{2}}$,
$\frac{CA}{CG}$ = $\frac{\sqrt{2}a}{2a}$ = $\frac{1}{\sqrt{2}}$,
∴$\frac{CF}{CA}$ = $\frac{CA}{CG}$ = $\frac{1}{\sqrt{2}}$.
∵∠ACF = ∠GCA,
∴△ACF∽△GCA.
(2) 解: ∵△ACF∽△GCA,
∴∠1 = ∠CAF.
∵∠CAF + ∠2 = ∠ACB = 45°,
∴∠1 + ∠2 = 45°.

答案：
(1) 证明: 设正方形的边长为a, 则
AC = $\sqrt{a^2 + a^2}$ = $\sqrt{2}a$.
∵$\frac{CF}{CA}$ = $\frac{a}{\sqrt{2}a}$ = $\frac{1}{\sqrt{2}}$,
$\frac{CA}{CG}$ = $\frac{\sqrt{2}a}{2a}$ = $\frac{1}{\sqrt{2}}$,
∴$\frac{CF}{CA}$ = $\frac{CA}{CG}$ = $\frac{1}{\sqrt{2}}$.
∵∠ACF = ∠GCA,
∴△ACF∽△GCA.
(2) 解:
∵△ACF∽△GCA,
∴∠1 = ∠CAF.
∵∠CAF + ∠2 = ∠ACB = 45°,
∴∠1 + ∠2 = 45°.

12. 如图,在$Rt△ABC$中,$∠ACB= 90^{\circ }$,以 AC 为直径的$\odot O$交 AB 于点 D,E 是边 BC 的中点,连接 DE.
(1)求证:DE 是$\odot O$的切线;
(2)若$AD= 4,BD= 9$,求$\odot O$的半径.

答案：

(1) 证明: 如图, 连接OD, CD,

∵∠ACB = 90°,
∴∠ACD + ∠DCB = 90°.
∵OC = OD,
∴∠OCD = ∠ODC.
∵AC是⊙O的直径,
∴∠ADC = 90°.
∴∠CDB = 180° - ∠ADC = 90°.
∵E是边BC的中点,
∴DE = CE = $\frac{1}{2}$BC.
∴∠DCE = ∠CDE.
∴∠ODC + ∠CDE = 90°.
∴∠ODE = 90°.
∵OD是⊙O的半径,
∴DE是⊙O的切线.
(2) 解:
∵AD = 4, BD = 9,
∴AB = AD + BD = 4 + 9 = 13.
∵∠ACB = ∠ADC = 90°,
∠CAB = ∠DAC,
∴△ACB∽△ADC.
∴$\frac{AC}{AD}$ = $\frac{AB}{AC}$.
∴AC² = AD·AB = 4 × 13
= 52.
∴AC = 2$\sqrt{13}$.
∴⊙O的半径为$\sqrt{13}$.