If two triangles have two sides equal to the middle line on the third side, are they congruent? The process of proof

Let the three sides corresponding to two triangles be A1, B1, C1; A1, B2, C2, and the middle line of the third side be LC1, LC2 respectively. If A1 = A2, B1 = B2, LC1 = LC2, then two triangles are congruent
Proof: we can know from the middle line theorem
Lc1=（2a1^2+2b1^2-c1^2)/4
Lc2=（2a2^2+2b2^2-c2^2)/4
And LC1 = LC2, i.e
2a1^2+2b1^2-c1^2=2a2^2+2b2^2-c2^2
And A1 = A2, B1 = B2
So C1 ^ 2 = C2 ^ 2, that is, C1 = C2
So two triangles are congruent

Proof: there are two sides and one side of the central line corresponding to the same two triangles congruent

As shown in the figure, in △ ABC and △ def, ab = De, BC = EF, an is the midline on BC, DM is the midline on EF, and an = DM. Prove: △ ABC ≌ Δ def. Prove: ? BC = EF, an is the midline on BC, DM is the midline on EF, ∥ BN = em, in △ ABN and △ DEM, ab = de & nbsp; BN = EM & nbsp

A trapezoid, with three lines, is divided into four shapes of the same size

What do you want to ask? Isosceles trapezoid, the bottom is twice the top

Among the 26 capital letters, several are axisymmetric

There are 16 of them

A cuboid wooden box has a volume of 180 cubic decimeters and a bottom area of 36 square decimeters. Its height is equal to the edge length of a cube
Surface area of

Height = 180 △ 36 = 5 decimeters
Cube surface area = 6 × 5 × 5 = 150 square decimeters

Given that the angle between vector a and B is π / 6, and the module of vector a = √ 3, the module of vector b = 1, find the cosine of the angle between vector a + B and a-b

|If the angle between vector a and vector B is w, then: |a-b |124;124;124;124;124;124;124;124;124;124;124;; ||124;|\124\\124\\\\\\124\\\\\\124\\\\\\\\\\\\\\\\\\35;178; - 2A * B + |b | & #178; = (√ 3) & #178; + 2 ×|a ×|b |× cos π / 6

In △ ABC, ab = BC = AC, D is the point on BC, de ⊥ AB is at point E, DF ⊥ AC is at point F

Equal
In RT △ BDE, ∠ B = 60 °, so BD = 2be; similarly, CD = 2cf
And ab = AC = BC = (BD + CD) = 2 (be + CF)
So AE + AF = AC + ab-be-cf = 3 (be + CF)
Δ AEF perimeter = AE + AF + EF = 3 (be + CF) + ef
Perimeter of quadrilateral ebcf = EB + BC + CF + EF = 3 (be + CF) + ef

How to find the tangent equation of a function when passing through a point not on the function image?

Let P (x0, Y0)
Tangent of function y = f (x) through p
Let the tangent point be (x, f (x))
From the slope relation
f'(x)=(f(x)-y0)/((x-x0)
We can get X
Solving tangent equation again

The relationship between translation vector and function translation
I don't quite understand the vector translation formula, so it's easy to get confused, so I want to know whether they can all use "up plus down minus", "left plus right minus"
Y = f (x) translation vector a = (h, K)
After translation
y=f(x-h)+k

You think:
(x, y) satisfies the functional relation y = f (x)
After the translation of vector a = (h, K)
X=x+h,Y=y+k
So the new y = f (x)
That is y + k = f (x + H)
So after translation, y = f (x + H) - K
As for whether it moves to the left or to the right
You can just draw the figure
After you remember this principle, don't be confused by adding left minus right

As shown in the figure, BP and CP divide the internal angle ∠ ABC and external angle ∠ ace of △ ABC equally to find the relationship between ∠ BPC and ∠ a

A: the reasons are as follows: according to the meaning of the title, let ∠ ABP = ∠ PBC = y, ∠ ACP = ∠ PCE = y ∵ ace be the outer angle of △ ABC, that is, ∠ a + 2Y = 2x & nbsp; ∠ a = 2x-2y = 2 (X-Y) ∵ PCE be the outer angle of △ PBC

If two triangles have two sides equal to the middle line on the third side, are they congruent? The process of proof