No one has been able to answer the triangle question so far, so let s try it!

I don't know your current academic qualifications, I hope you can understand the following answers, in fact, this is an inductive summary of the question) The answers are as follows:

At one point: can be divided into triangles 2+1=3

At two points: 3+2+1=6

At three points: 4+3+2+1=10

Then n points: (n+1)+n+(n-1)+3+2+1 is the sum of an equal difference series, which is easy to obtain using the summation formula:

n+1）+n+（n-1）+.3+2+1=(n+2)(n+1) 2 means that when there are n points on the edge of bc, the triangle can be divided into =(n+2)(n+1) 2 triangles.

Small triangles = n.

A triangle consisting of two small triangles = n-1.

And so on. A triangle consisting of n small triangles = 1.

Total = n+(n-1)+1=n*(n+1)/2

One point d, 3=(1+1)*(1+2) 2Two points d,e, 6=(2+1)*(2+2) 2 three points: 10=(3+1)*(3+2) 2n points have (n+1)(n+2) 2 triangles.

At one point: can be divided into triangles 2+1=3

At two points: 3+2+1=6

Then n points: (n+1)+n+(n-1)+3+2+1 is the sum of an equal difference series, which is easy to obtain using the summation formula:

n+1）+n+（n-1）+.3+2+1=(n+2)(n+1) 2 is divided into (n+2)(n+1) 2

triangles.

I knew this question in elementary school, and the answer is: (n+2)(n+1) 2, there are 1, 2, 3, 4 points, 3, 6, lo, 15

Solution 1:The quadrilateral aobf is a parallelogram, ah=bh, that is, h is the midpoint of ab, and g is the midpoint of ab, and the point g, the point h coincides, and f,h,o collinear, f,g,o collinear, and c,o,g collinear, f,g,o,c collinear,FO, CO collinear.

AF OB, AF OD, D is the AC midpoint, and OD is the ACF mediano is the midpoint of cf, i.e., fo=co

Solution 2:o is the center of gravity of abc (the intersection of the three midlines), ob=2od, (here is the conclusion).

The quadrilateral aobf is a parallelogram, af=ob, af=2ob, d is the ac midpoint, and od is the acf mediano is the midpoint of cf, that is, fo=co, and fo,co are collearlinear.

It's not easy to code words, hope!

Solution 1: As shown in the figure, let the line segment fo and the line segment ab intersect at the point hThe quadrilateral aobf is a parallelogram, ah=bh, i.e. h is the midpoint of ab, and g is the midpoint of ab, and g, g, h....

The solution consists of three sides of a triangle that are 20 centimeters, 10 centimeters, and 14 centimeters long, and the two such triangles are assembled into a parallelogram, and when the sides of the parallelogram are 20 and 10 long, the circumference of the parallelogram is at most 2*(20+10)=60 centimeters.

The sides of the three triangles of 20cm, 10cm and 14cm are respectively regarded as the diagonals of the parallelogram, and the other two sides are the two sides of the parallelogram, such as 20cm as the diagonal of the parallelogram, the sides of 10cm and 14cm become the two sides of the parallelogram, and the perimeter is (10+14)*2=48, and the others are the same, the maximum circumference should be 68cm

The three cases are 60, 48 and 68

The radius of the inscribed circle is 3

I'll draw a picture for you.

a+c=24

b+c=25 (use the right-angled edge to find the hypotenuse).

This gives a=3

The radius of the inscribed circle r=3, and the formula r= twice the area of the right triangle and the circumference of the right triangle (r=2s l).

or r = product of two right-angled edges of a right-angled triangle Perimeter of a right-angled triangle (r=a*b l).

2.In the quadrilateral defb, because de bc, ef ab, defb is a parallelogram, so db=ef

3.Because d is the midpoint of ab, ad=db and ad=ef because db=ef

4.In AED and ECF.

Because ab ef so a= fec

de bc so aed= c

AD=EF (verified).

So ade efc

So ae=ec

The three sides of the triangle are a, b, c (you can draw your own drawing), if you take a midpoint from the A side and draw a line L through this midpoint, this line is parallel to the B side, then the line L and the C side must intersect at a point, and this point must also be the midpoint of the C side, this line is called the middle line of the C side.

To put it simply, you find that after making parallel lines, a small triangle appears, and the angle size of this small triangle and the original triangle have not changed! It's the equivalent of amp scaling! You can draw a few triangles, not necessarily in the middle of one side, but also in a third of an edge and draw parallel lines!

Look at the relationship between the two triangles!

It is the "median line" of the triangle.

It refers to the median line of the triangle.

Median line, see median line theorem.