The side length of square ABCD is 12, e is a point on BC and be = 5, the vertical bisector Mn of AE intersects CD at point m and ab at point n, and the length of Mn is calculated

Do NF ∥ BC hand CD to f
Then NF = BC = ab
∠ FMN = ∠ anm (internal stagger angle) = ∠ bea (congruence)
∴Rt△ABE≌Rt△NFM （AAS)
So me = AE = √ (AB & # 178; + be & # 178;) = 13

The side length of square ABCD is 2, AE = EB, the two ends of line Mn slide on CB and CD respectively, and Mn = 1
When cm is the value, the triangle AED is similar to the triangle with m, N, C as the vertex
If you answer right, you will be rewarded

1. When AE corresponds to cm, cm: AE = Mn: De can be obtained from triangle similarity, and de = 5 / 5 under root can be obtained from Pythagorean theorem, so cm = AE * Mn / de = 5 / 5 under root
When cm and AD are paired, cm: ad = Mn: De can be obtained from triangle similarity, and de = 5 / 5 under root can be obtained from Pythagorean theorem, so cm = AE * Mn / de = 5 / 5 under 2 * root

Fold the square paper ABCD with side length of 12 so that point a falls on point E on side CD, and the crease is Mn. If the length of Mn is 13, what is the length of CE?
Urgent............................ Ladies and gentlemen, handsome boys and beauties, please help me ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~!!!!!!!!!!!!!!!!!!!!!!!!!!!!

CE=7
It can be found that de = 5
It's not easy to explain if there's no graph. It's OK to make a parallel line of AB through M or n
It can be proved that triangles are congruent, 5,12,13 are trilateral

Fold the square ABCD in half to get the crease Mn (as shown in Fig. 1). After unfolding, fold the square ABCD along CE so that the point B 'falls on the point B' on Mn and connects B'd (as shown in Fig. 2). Try to find the degree of ∠ BCB 'and ∠ ADB'

∵ point B falls at point B ′ on Mn, and the square ABCD is folded in half to get creases Mn, ∵ BC = B ′ C, BB ′ = B ′ C, ∵ BC = BB ′ = B ′ C, ∵ B ′ BC is equilateral triangle, ∵ BCB ′ = 60 °, ∵ B ′ CD = 30 °, ∵ DC = B ′ C, ∵ CB ′ d = ∵ CDB ′, ∵ CB ′ d = ∵ CDB ′ = 12 × 150 ° = 75 ° and ∵ ADB ′ = 15 °

ABCD is a square and E is a point on BC. Fold the square in half so that a and e coincide and the crease is Mn. If Tan angle AEN is 1 / 3, DC + CE = 10

Mn is AE vertical bisector
Make MF ⊥ AB in F
It can be proved that ∠ FMN = ∠ BAE = ∠ NEA
tan∠EAB=1/3
EB=BC/3
CE=2BC/3
DC+CE=5BC/3=10
Side length = 6

Let AB = 2, when CECD = 12, AmBn = 1515. If CECD = 1n (n is an integer), AmBn = (n − 1) 2n2 + 1 (n − 1) 2n2 + 1. (expressed by a formula containing n)

Given that CECD = 1n (n is an integer), and CD = 2, then CE = 2n, de = 2n − 2n; let am = a, BN = B; in RT △ nce, NE = BN = B, NC = 2-B, from Pythagorean theorem, we get: NE2 = NC2 + CE2, that is, B2 = (2-B) 2 + (2n) 2; the solution is: B = N2 + 1n2, BN = ne = N2 + 1n2, NC = 2-B = N2 − 1n2

As shown in the figure, given that e is a point on the edge CD of square ABCD and Ce: de = 1:2, the length of AB is a nm ⊥ be, then Mn is long
Figure: A M D
E
F
B N C
(square ABCD, m on ad, E on CD, N on BC, connecting Mn and be at point F)

If CP ‖ nm. P ∈ ad, then ⊿ CPN ≌ ⊿ bec (AAS),
MN＝PC＝BE＝（√（1+1/9））a＝√10a/3≈1.0541a.

As shown in the figure, in square ABCD, CE ⊥ Mn, if ∠ MCE = 35 °, then the degree of ∠ anm is______ ．

If NP ⊥ BC is made to P through N, then NP = DC, ∫ MCE + ∠ NMC = 90 °, MNP + ∠ NMC = 90 ° and ∪ MCE = ∠ MNP, ∪ MNP = ∠ mcenp = CB ∠ NPM = ∠ CBE, ≌ BEC ≌ PMN, ∪ MCE = ∠ PNM and ∪ anm = 90 ° - MCE = 55 ° in △ MnP and △ ECB

In square ABCD, points e, m and N are on the sides of AB, BC and ad respectively, CE = Mn, and angle MCE = 35 degrees

GM is perpendicular to AD and G is perpendicular to foot
A quadrilateral is an ABCD, a square
That is to say, △ ECB is a right triangle (all four corners of a square are right angles)
∵MG⊥AD
That is to say, △ NMG is a right triangle
A quadrilateral is an ABCD, a square
⊥ BA ⊥ ad (the adjacent sides of a square are perpendicular to each other)
∵AD⊥GM AB⊥AD
{AB / / GM (if both lines are perpendicular to the third line, then the two lines are parallel to each other)
A quadrilateral is an ABCD, a square
‖ AD / / BC (parallel to the opposite side of a parallelogram)
∵AB//GM AD//BC
{AB = GM (the parallel line segments between two parallel lines are equal)
A quadrilateral is an ABCD, a square
｜ AB = BC (the four sides of a square are equal)
∵AB=GM AB=BC
∴BC=GM
∵ △ NMG is a right triangle △ ECB is a right triangle BC = GM CE = Mn
∴△NMG≌△ECB （HL）
The corresponding angles of congruent triangles are equal
∵∠MCE=35°
∴∠BEC=55°
∵∠BEC=55°∠ANM=∠BEC
∴∠ANM=55°

The side length of square ABCD is 12, e is a point on BC and be = 5, the vertical bisector Mn of AE intersects CD at point m and ab at point n, and the length of Mn is calculated