Here are two kinds of certificates for you, please help yourself. I'm sorry that the vector symbol can't be solved, which may bring inconvenience to your reading. )

Proof method 1

Make a picture first, make a b,

The two median lines of c intersect with ac in M and ab in N respectively, so M and N are the midpoint of AC and AB. Connect mn.

Let vector bp=λ vector pm, vector cp=μ vector pn(λ, μ is a real number not equal to 0).

Vector bc= vector pc- vector pb= vector bp- vector cp=λ vector pm-μ vector pn,

Vector nm= vector pm- vector pn, vector bc=2 vector nm.

Therefore, λ vector pm-μ vector pn=2 vector pm-2 vector pn.

That is, (λ-2) vector pm-(μ-2) vector pn=o vector.

Because the vector pm and the vector pn are not linear, λ = 2 and μ = 2.

So vector bp=2 vector pm.

It is proved that the intersection of two median lines divides bm into 2: 1. Similarly, it can be proved that the intersection of another midline and bm also has this property, so the three midlines of the triangle intersect at one point and divide each line equally at the ratio of 1: 2.

Get a license.

Proof method 2

Make a triangle ABc, let d, e, f E and f be the midpoint of bc, CA and ab respectively, take any point o on the plane, let vector oa=a, vector ob=b and vector oc = C.

Then the vector od= 1/2(b+c), the vector of= 1/2(a+b) and the vector oe= 1/2(c+a).

Let p be the bisector on ad and satisfy vector ap=2 vector pd.

Then the vector op= 1/3, and the vector oa+2/3od= 1/2a+2/3.

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1/2(a+b)= 1/3(a+b+c)

Similarly, P is also the bisector of be and cf, so the three median lines intersect at point P..

The three median lines of a triangle intersect at one point and divide each line equally. The ratio is 1: 2.