The key to this problem is to make the correct auxiliary line. In fact, the second question is a right triangle with a center line, but it is very easy.

The first question goes without saying.

(2) 1。 There is k = 3, so ∠EFD = K∠AEF. The reason for this is the following：

Connect CF and the local location g of extension line to expand broadcast management,

F is the midpoint of AD, ∴AF = FD.

In parallelogram abcd, ABCD, ∴∠G =∠DCF.

At △AFG, △CFD,

∠∠G =∠DCF，∠G =∠DCF，AF = FD，

∴△AFG≌△CFD(AAS)。 ∴CF = GF，AG = CD .

∵CE⊥AB, f is the GC side of midpoint∴ ∴EF = GF. ∴∠AEF =∠G .

∵AB = 5, BC = 10, F is the midpoint of AD, ∴AG = 5, AF = AD = BC = 5, ∴AG = AF.

∴∠AFG =∠G .

In △AFG, ∠EFC =∠AEF +∠G = 2∠AEF,

And ≈CFD =∠afg, ∴∠CFD =∠ AEF.

∴∠efd =∠EFC+∠CFD = 2∠AEF+∠AEF = 3∠AEF，

Therefore, there is a positive integer k = 3, so ∠EFD = 3∠AEF.

2。 At BE = X, ag = cd = ab = 5, ∴EG = AE+AG = 5-X +5 = 10-X,

CE2 = BC2-BE2 = 100-X2 when the retention time is △BCE.

In △CEG in RT, CG2 = EG2+Ce2 = (10-x) 2+100x2 = 200-20x.

∫cf = gf(① authentication), ∴CF 2 =(CG2)/ 4 = 50-5X.

∴ce2-cf 2 = 100-x2-50+5x =-x2+5x+50 =-(x-2.5)2+50+(2.5)2

When x = 2.5, this is one point. When e is the midpoint of the maximum value of AB, CE2-CF2. Tan ∠DCF =∠ Tan BGC = EC ratio GE/& gt；; Main 3

& ltBr =( 15) Knowledge test also depends on the brain. Don't be afraid of this topic. What is the result of the topic?