答案：9. D
∵AE = BE,DF = FB,
∴EF是△BAD的中位线,
∴EF//AD,AD = 2EF,
∵EF = 2,
∴AD = 4。
∵DF⊥CE,
∴∠BFE = ∠BFC = 90°,
∵tan∠ABD = $\frac{1}{4}$,
∴$\frac{EF}{BF}$ = $\frac{1}{4}$,
∵EF = 2,
∴FB = 8,
∵AF//DC,EF//AD,
∴四边形ADCF是平行四边形，
∴AD = CF = 4,
∴BC = $\sqrt{CF^{2} + BF^{2}}$ = $\sqrt{4^{2} + 8^{2}}$ = 4$\sqrt{5}$。故选D。

答案：
10. C　如图,过点C作CD⊥AB,垂足为D,
∵EF⊥AB,
∴CD//EF,
∴∠DCE = ∠CEF,
∴在Rt△CDE中,sin∠DCE = sin∠CEF = $\frac{DE}{CE}$ = $\frac{3}{5}$,
∴设DE = 3x,则CE = 5x,
∴CD = $\sqrt{CE^{2} - DE^{2}}$ = 4x,
∵∠ACB = 90°,CE是斜边AB上的中线，
∴CE = BE = EA = 5x,
∴AB = 2BE = 10x,BD = BE - DE = 2x,
∵在Rt△BCD中,BC² = CD² + BD²,BC = 4,
∴4² = (4x)² + (2x)²,
∴x = $\frac{2\sqrt{5}}{5}$(舍负)，
∵CD//EF,
∴△ACD∽△AFE,
∴$\frac{EF}{CD}$ = $\frac{AE}{AD}$,
∴$\frac{EF}{4x}$ = $\frac{5x}{5x + 3x}$,
∴EF = $\frac{5}{2}$x = $\frac{5}{2}$×$\frac{2\sqrt{5}}{5}$ = $\sqrt{5}$,
∵AE = 5x = 2$\sqrt{5}$
∴S△AEF = $\frac{1}{2}$EF·AE = $\frac{1}{2}$×$\sqrt{5}$×2$\sqrt{5}$ = 5。故选C。

答案：
11. 答案　$\frac{24}{5}$ ≤ BP ≤ 6
解析　如图,过点A作BC的垂线，垂足为M,在Rt△ABM中,cos∠ABM = $\frac{BM}{AB}$,
∴BM = $\frac{3}{5}$×5 = 3,
∴AM = $\sqrt{5^{2} - 3^{2}}$ = 4。
∵AB = AC,
∴BC = 2BM = 6。过点B 作AC的垂线,垂足为N,
∵S△ABC = $\frac{1}{2}$BC·AM = $\frac{1}{2}$AC·BN,
∴BN = $\frac{BC·AM}{AC}$ = $\frac{6×4}{5}$ = $\frac{24}{5}$,即当BP⊥AC时，BP取得最小值,为$\frac{24}{5}$。当点P与点C重合时,BP取得最大值，为6,
∴线段BP长的取值范围是$\frac{24}{5}$ ≤ BP ≤ 6。

答案：
12. 答案　$\sqrt{5}$
解析　如图,过点A作CE边的垂线，垂足为M,
∵AM⊥CE,∠ACB = 90°,
∴∠ACB = ∠AMC。
∵CE//BD,
∴∠BDC = ∠ACM。
在△BCD和△AMC中，$\begin{cases} ∠ BCD = ∠ AMC, \\ ∠ BDC = ∠ ACM, \\ BD = AC, \end{cases}$
∴△BCD≌△AMC(AAS),
∴AM = BC = 2。
在Rt△AME中,tanE = $\frac{AM}{ME}$,
∴ME = $\frac{2}{2}$ = 1,
∴AE = $\sqrt{1^{2} + 2^{2}}$ = $\sqrt{5}$

答案：
13. 解析　(1)如图所示。
(2)如图,连接OP,则OP = OA,
∴∠OAP = ∠OPA。
∵AC = BC,
∴∠CAB = ∠CBA,
∴∠OPA = ∠CBA,
∴OP//BC,
∵PQ⊥BC,
∴OP⊥PQ,
∵OP为⊙O的半径，
∴PQ为⊙O的切线。
(3)如图,连接CP,
∵AC是⊙O的直径，
∴∠APC = 90° = ∠BPC,
又
∵AC = BC,
∴AP = PB = $\frac{1}{2}$AB = 7,
∵∠PQB = ∠BPC,∠PBQ = ∠CBP,
∴∠BPQ = ∠BCP,
在Rt△CBP中,tan∠BCP = $\frac{BP}{CP}$ = $\frac{7}{\sqrt{25^{2} - 7^{2}}}$ = $\frac{7}{24}$,
∴∠BPQ的正切值为$\frac{7}{24}$。

答案：
14. 答案　2$\sqrt{6}$ + 3
解析　如图,过点A作AH⊥BC于H,过点C作CE⊥AD于E,连接AC。在Rt△ABH中,tanB = $\frac{AH}{BH}$ = $\frac{3}{4}$,
∴设AH = 3k,则BH = 4k,
∴AB = $\sqrt{AH^{2} + BH^{2}}$ = 5k,
∵AB = 10,
∴k = 2,
∴AH = 6,BH = 8,
∵BC = 10,
∴CH = BC - BH = 10 - 8 = 2,
∴AC = $\sqrt{AH^{2} + CH^{2}}$ = $\sqrt{6^{2} + 2^{2}}$ = 2$\sqrt{10}$,
∵∠D + ∠ECD = 90°,∠B + ∠D = 90°,
∴∠ECD = ∠B,
∴tan∠ECD = tanB = $\frac{3}{4}$,
∴在Rt△CED中,tan∠ECD = $\frac{DE}{CE}$ = $\frac{3}{4}$,
∴设DE = 3m,则CE = 4m,
∵CD = 5,
∴(3m)² + (4m)² = 5²,
∴m = 1(负值已舍),
∴DE = 3,CE = 4,
∴AE = $\sqrt{AC^{2} - CE^{2}}$ = $\sqrt{(2\sqrt{10})^{2} - 4^{2}}$ = 2$\sqrt{6}$,
∴AD = AE + DE = 2$\sqrt{6}$ + 3。