In a triangle, the line segment connecting the vertex and the midpoint of the opposite side is called the center line of the triangle.

Any triangle has three median lines, which are all inside the triangle and intersect at one point.

By definition, the center line of a triangle is a line segment.

Because a triangle has three sides, it has three median lines.

The three median lines intersect at one point. This point is called the center of gravity of the triangle.

The area of two triangles divided by the center line of each triangle is equal.

2 attributes

Let the opposite sides of angles A, B and C of ⊿ABC be A, B and C respectively.

1, the three midlines of the triangle are all within the triangle.

2. The length of the three midlines of the triangle:

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ma=( 1/2)√2b？ +2c？ -a？ ;

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mb=( 1/2)√2c？ +2a？ -B? ;

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mc=( 1/2)√2a？ +2b？ -c？ .

(Ma, MB and MC are the lengths of the midline subtended by angles A, B and C, respectively)

3. The three center lines of a triangle intersect at a point, which is called the center of gravity of the triangle.

4. The midline of hypotenuse of right triangle is equal to half of hypotenuse.

5. The triangle area formed by the center line of the triangle is equal to 3/4 of the triangle area.

3 midline theorem

The midline theorem (pappus theorem), also known as Apollonius theorem, is a theorem in Euclidean geometry, which expresses the relationship between the lengths of three sides of a triangle and the midline.

Theorem content: The sum of the squares of the opposite sides of the center line of a triangle is equal to half the square of the bottom line and twice the sum of the square of the center line.

That is, for any triangle △ABC, let I be the midpoint of BC line and AI be the center line, there is the following relationship:

AB？ +AC？ =2(BI？ +AI？ )

Or AB? +AC？ = 1/2BC？ +2AI？

4 Theorem Proof

In addition to the methods given in the figure, c 1c2clone gives two other conventional proof methods here:

One is to establish a coordinate system in horizontal and vertical directions with the midpoint as the origin.

Let: A(m, n), B(-a, 0), C(a, 0),

Then: (AD)? +(CD)？ =m？ +n？ +a？

(AB)？ +(AC)？ =(m+a)？ +n？ +(m-a)？ +n？ =2(m？ +a？ +n？ )

∴(AB)？ +(AC)？ =2[(AD)？ +(CD)？ ]

The second is to use the cosine theorem twice for the same angle in different triangles, for example, to use the cosine theorem for ∠B (or ∠C) in △ABD and △ABC (or △ACD and △ABC) in the diagram, so as to directly obtain the relationship between the sides of the triangle, and then prove it.