The area of a triangle is 48cm2, and the area of the triangle surrounded by its three median lines is

The area of a triangle is 48cm2, and the area of the triangle surrounded by its three median lines is

The perimeter of a midline triangle is half that of the original triangle
So the area is a quarter of the original triangle
So the area is 12cm ²

Find the area ratio of the triangle surrounded by the three median lines of the triangle to the original triangle

1/4
The two triangles are similar triangles, and the similarity ratio is 1:2
The square of the area ratio equal to the similarity ratio is 1 / 4

Prove that the three midlines of a triangle intersect at one point

prove:
It is known that in △ ABC, BD is the AC midline, CE is the AB midline, and BD and CE intersect at point o,
Verify the midline AF crossing point O of BC
Extend Ao to BC at f '
Make BG parallel EC cross Ao extension line at G
Then, since e is the midpoint of AB, O is the midpoint of Ag
If GC is connected, OD is the median line in triangular AGC
BD parallel GC
So BOCG is a parallelogram
F 'bisect BC
F 'coincides with F
Central line AF of BC crosses point o
Therefore, it can be proved that the three midlines of the triangle intersect at one point
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As shown in the figure, if you know △ ABC, please make a triangle so that its area is twice the area of △ ABC

As shown in the figure, extend BC to d so that CD = BC,
Then BD = 2BC,
Set the distance from point a to BC as h,
Then s △ abd = 1
2BD•h=1
2•2BC•h=2•1
2•BC•h=2S△ABC，
Therefore, △ abd is the triangle

As shown in the figure, in △ ABC, BD = 2da, CE = 2EB, AF = 2FC, then the area of △ ABC is the area of shadow triangle______ Times

Connecting De, DF and EF, from BD = 2da, CE = 2EB and AF = 2FC, it can be seen that triangle ADF area = triangle BDE area = triangle CEF area = 13 × 23 triangle ABC area = 29 of triangle ABC area according to the swallow tail theorem: the def area of the middle triangle is 13 of the ABC area of the triangle, and the shadow area = the def area of the triangle

As shown in the figure, D. e is the two points in the triangle ABC, connecting dB, de and CE. How about the size relationship between ab + AC and BD + de + EC,

Extend CE and AB to M
Extend BD and cm to F
In △ ACM
AC+AM＞MF+FE+CE（1）
In △ BMF
BM+MF＞BD+DF（2）
In △ FDE
DF+FE＞DE（3）
（1）+（2）+（3）
AC+AM+BM+MF+DF+FE＞MF+FE+CE+BD+DF+DE
AC+AB＞BD+DE+EC