CB = CB' is obtained by folding in half. In RT△B'FC, SIN∠CB'F =
CF/CB′？ = 1/2 ? , get ∠CB'f = 30,
The second question, the connecting line BB ′ and CG intersect at point K, from folding, we know that ∠ B ′ AE = ∠ B ′ Be, from ∠ B ′ Be+∠ KBC = 90, ∠ KBC+∠ GCB = 90, we get ∠.
Solution: (1) As shown in figure 1, it can be seen from folding that ∠ EFC = 90, CF= 1/2CD, ∵ quadrilateral ABCD is a square.
∴CD=CB, this is a detailed answer. /exercise/math/7989 14 quadrilateral ABCD is a square piece of paper. First, fold the square ABCD in half, so that BC and AD overlap, and the crease is EF. Then flatten the square, and then fold it along the straight line CG, so that the point B falls on EF, and the corresponding point is B'.
Mathematical thinking: (1) the number of times to find ∠CB'F; (2) As shown in Figure 2, on the basis of Figure 1, connect AB', try to judge the size relationship between ∠B'AE and ∠GCB', and explain the reasons;
The topic is really long. Seeing such a topic will make it impossible to do what you can, so it is very important to adjust your mentality during the exam. I believe you will understand after reading the answer. If you don't understand, you can keep asking me. Come on ~ I hope to adopt it if it is useful.