How to make a 30 degree angle when the rectangular paper is folded twice? It can only be folded twice

Use a short side of a piece of paper inward, and the long side of the same angle is close to the folded hypotenuse. The creases of the equal length side and the short side coincide. After the creases of the short side and the long side coincide, compact the fold line, and you will have an angle of 30 degrees! The simple thing is to divide a right angle into three equal parts!

As shown in the figure, fold the two corners of the rectangle ABCD, and the creases are EF and Hg, so that HD and BF are on the same straight line. It is known that the two opposite sides of the rectangle are respectively parallel. Try to explain that the two creases are also parallel to each other

∵ ad ∥ BC, ∥ DHF = ∥ BFH, known from folding: ∥ FHG = ∥ DHG = 12 ∥ DHF ∥ HFE = ∥ BFE = 12 ∥ BFH, ∥ FHG = ∥ HFE, ∥ EF ∥ Hg, that is, the two creases are parallel to each other

There is a rectangular paper ABCD, ab = 6cm, BC = 8cm. Fold the paper ABCD along EF, and point B coincides with point d to find the crease
Finding the length of crease ef

The first step is to determine the crease
In rectangular ABCD, connect BD and make the vertical bisector of BD, intersect AD and BC with points E and f respectively. The line EF is the crease
The second step is to find the length of EF
Connect be and FD, let EF and BD intersect at point G
Easy to get: EF and BD are equally divided vertically
The bfde of quadrilateral is rhombic
Let BF = x, then CF = 8-x, DF = X
In RT △ CDF, CF ^ 2 + CD ^ 2 = DF ^ 2
Namely: (8-x) ^ 2 + 6 ^ 2 = x ^ 2
The solution is: x = 25 / 4
Easy to get: diagonal BD = 10
∴BG=5.
In RT △ BGF, BG ^ 2 + GF ^ 2 = BF ^ 2
Namely: 5 ^ 2 + GF ^ 2 = (25 / 4) ^ 2
The solution is GF = 15 / 4
So EF = GF * 2 = 15 / 4 * 2 = 15 / 2
A: the length of the crease is 15 / 2

How to make a 30 degree angle when the rectangular paper is folded twice? It can only be folded twice