10. (2025·编写)如图,已知Rt△ABC∽Rt△CAD,若AC= 3,BC= 4,求AD的长.

[解]∵AC = 3,BC = 4,∠ACB = 90°,∴由勾股定理得AB = 5.
∵Rt△ABC∽Rt△CAD,∴$\frac{BC}{AD}$=$\frac{AB}{AC}$,即$\frac{4}{AD}$=$\frac{5}{3}$,解得AD = $\frac{12}{5}$.
∴AD的长为$\frac{12}{5}$.

11. (2025·编写)如图,△ABC∽△ADE,BC的延长线过点E,∠ACB= ∠AED= 105°,∠CAD= 10°,∠B= 50°,则∠DEF的度数是______.

35°

12. (2025·编写)如图,在△ABC中,若DE//BC,EF//CD,AE= 2EC,则AF:FD:DB= ______.

4:2:3

13. (2025·编写)如图,在△ABC中,AB= 6,AC= 4,∠A= 90°,D是AB边的中点,点E在直线AC上,且△ADE与△ABC相似,则CE= ______.

2或0.5或6或8.5

14. (2025·编写)如图,延长正方形ABCD的一边CB至点E,ED与AB相交于点F,过点F作FG//BE交AE于点G.求证:GF= FB.

[证明]∵四边形ABCD为正方形,∴BF//CD,∴$\frac{BF}{CD}$=$\frac{EF}{ED}$.
∵FG//BE,∴GF//AD,
∴$\frac{GF}{AD}$=$\frac{EF}{ED}$,
∴$\frac{GF}{AD}$=$\frac{BF}{CD}$.
∵AD = CD,∴GF = BF;

15. (2025·编写)如图,在三角形ABC中,D为BC的中点,AF= 2BF,CE= 3AE,连接CF交DE于点P.求$\frac{EP}{DP}$的值.

[解]如图,过点E作EG//CB交AB于点G,交CF的延长线于点H.

∵EG//CB,CE = 3AE,
∴$\frac{EG}{BC}$=$\frac{AE}{AC}$=$\frac{AG}{AB}$=$\frac{1}{4}$,
∴可以设EG = m,则BC = 4m,CD = 2m.
∵AF = 2BF,∴设BF = a,则AF = 2a,AB = 3a.
∴AG = $\frac{1}{4}$AB = $\frac{3}{4}$a,FG = AF - AG = 2a - $\frac{3}{4}$a = $\frac{5}{4}$a.
∵GH//BC,∴$\frac{GH}{BC}$=$\frac{FG}{BF}$,∴$\frac{GH}{4m}$=$\frac{\frac{5}{4}a}{a}$,∴HG = 5m,∴EH = 6m.
∵EH//CD,∴$\frac{EP}{PD}$=$\frac{EH}{CD}$=$\frac{6m}{2m}$ = 3.