As shown in the figure, there is a quadrilateral piece of paper ABCD, ab ‖ CD, ad ‖ BC, ∠ a = 60 degrees. Fold the paper along the crease Mn and PQ respectively, so that point a coincides with point E on the edge of AB, point C coincides with point F on the edge of CD, eg bisection ∠ MEB intersects CD at g, FH bisection ∠ PFD intersects AB at h. try to explain: (1) eg ‖ FH; (2) me ‖ PF

(1) ∵ point a is folded along Mn to coincide with point E, point C is folded along PQ to coincide with point F, ∵ mea = ∵ a, ∵ PFC = ∵ C, (1 point) ∵ ab ∥ CD (known), ∵ a = 60 °, ∵ D + ∵ a = 180 ° (two lines are parallel and complementary to each other's inner angle), ∵ d = 120 °, ∵ ad ∥ BC (known), ∵ C + ∵ d = 180 ° (two lines are parallel and complementary to each other's inner angle), ∵ C = 60 °, ∵ mea = ∵ PFC= There are three parts: (3 points) DC \ (3 points) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(2)

In rectangular paper ABCD, ab = 3, ad = 5. As shown in the figure, fold the paper so that point a falls at a 'on the edge of BC, and the crease is PQ. When point a' moves on the edge of BC, the end point P.Q of the crease also moves. If the limiting points P and Q move on the edge of AB and ad respectively, the maximum distance that point a 'can move on the edge of BC is ()
A. 1B. 2C. 3D. 4

As shown in Figure 1, when point d coincides with point Q, according to the folding symmetry, we can get a ′ d = ad = 5. In RT △ a ′ CD, a ′ D2 = a ′ C2 + CD2, that is 52 = (5-a ′ b) 2 + 32, we can get a ′ B = 1. As shown in Figure 2, when point P coincides with point B, according to the folding symmetry, we can get a ′ B = AB = 3, ∵ 3-1 = 2, and the maximum distance that a ′ can move on the edge of BC is 2

Rectangular paper ABCD, ad = 9, ab = 12, fold the paper so that a and C coincide. What is the broken line?

AC = 15, let the midpoint of AC be e, the broken line intersects AB at F, (the broken line bisects the diagonal AC vertically)
AE=7.5
Triangle AEF is similar to ABC,
EF/AE=BC/AB=9/12,
EF=22.5/4
Broken line length = 2ef = 11.25

There is a rectangular piece of paper ABCD. It is known that ab = 2 and ad = 5. Fold this piece of paper so that point a falls at point E on edge BC, and the crease is Mn. Mn intersects AB at M and ad at n (1) (2) if be = 2, try to find the length of am. (3) when point e moves on BC, let be = x, an = y, try to find the analytic expression of Y with respect to x, and write out the value range of X. (4) connecting De, is there such a point e that △ ame is similar to △ dne? If it exists, request the length of be. If not, please explain the reason

(1) As shown in the figure: (3 points) (2) connect me, as shown in Figure 1, ∵ be = 2, let BM = x, then me = 2-x. according to Pythagorean theorem, we can get: BM2 + be2 = me2, ∵ 2 + x2 = (2-x) 2, ∵ 2 + X2 = 4-4x + X2, ∵ x = 12, ∵ am = 32; (3) extend nm intersection CB

As shown in the figure, the bottom of p-abcd is square, PA is vertical to ABCD, and PA = ad, e is a point on PC, PD is vertical to Abe
Verify that e is the midpoint of PC

Set up the space coordinate system to solve. Take a point as the vertex, ad as the X axis, AB as the Y axis, AP as the Z axis. Let the side length of the square be 1, then the points are a (0,0,0), B (0,1,0), D (1,0,0), C (1,1,0), P (0,0,1). Let e (x, y, z)
The vectors are PD = (1,0, - 1), AE = (x, y, z), be = (x, Y-1, z), PE = (x, y, Z-1), PC = (1,1, - 1)
Because PD is perpendicular to Abe, there are vectors PD × AE = 0, PD × be = 0, and x-z = 0, ①
Because e is on PC, there is PE = MPC, that is, x = m, y = m, Z-1 = - M
We can get the solution: x = y = z = 0.5
So e (0.5,0.5,0.5), so e is the midpoint of PC

As shown in the figure, take any point m on the side BC of the square ABCD, pass through point C to make cn ⊥ DM, and cross AB to N. let the intersection of the diagonal of the square be o. try to determine the relationship between OM and on, and explain the reason

In △ DCM and △ CBN, ∵ NCM ≌ CBN (ASA), ∵ cm = BN

As shown in the figure, in the parallelogram ABCD, AB is equal to 2ad, and M is the midpoint of AB, connecting DM and MC. What's the position relationship between DM and MC

It is proved that ab = 2ad, ab = 2am
∴AD=AM,∠AMD=∠ADM
AB‖CD,∴∠AMD=∠CDM
So DM is the bisector of ADC
Similarly, CM is the bisector of BCD
∵∠ADC+∠BCD=180,∴∠MDC+∠MCD=90
∴DM⊥CM

In trapezoidal ABCD, AD / / BC, M is the midpoint of lumbar AB, and AD + BC = DC

Extending DM to BC in n
All △ AMD is equal to △ BNM
MD＝MN
NC＝NB＋BC＝AD＋BC＝DC
The △ NCD is isosceles
Cm is the bottom line, and the original proposition is proved

As shown in the figure, in trapezoidal ABCD, it is known that ab ‖ CD, M is the key point of AB, and DM = cm. Try to explain the reason why trapezoidal ABCD is isosceles trapezoid

It is proved that the ∵ quadrilateral ABCD is a parallelogram
∴∠MDC=∠DMA ∠MCD=∠CMB
And ∵ CN = DM
The Δ MDC is an isosceles triangle
∴∠MDC=∠MCD
∴∠DMA =∠CMB
And ∵ m is the midpoint of ab
∴AM=BM
According to [edge] or [SAS]
It is known that △ AMD is equivalent to △ BMC
∴AD=BC
The trapezoid ABCD is isosceles trapezoid
Enough detail

As shown in the figure: in square ABCD, the crossing point D is DP, AC is crossed at point m, AB is crossed at point n, and the extension line of CB is crossed at point P. if Mn = 1, PN = 3, the length of DM is______ ．

Let DM = x, Pb = y, and the side length of square ABCD be Z, then DN = x + 1 ∵ ad ∥ PC ∥ ADM ∥ CPM, △ PNB ∥ DNA ∥ pndn = PBAD ∥ 3x + 1 = YZ, 4x = y + ZZ = YZ + 1, 3x + 1 = 4x-1, 3x = 4 (x + 1) - x2-x, x = 2 or - 2 (rounding off), ∥ x = 2