Can anyone help to prove why the area of the median triangle in the inner triangle of the circle is

Sorry, because I'm lazy, I'll dictate it directly) First make a triangle in the circle.

The intersection of the triangle and the circle is set to a, b, c and the center of the circle is set to o to connect ao, bo, co

So there are three triangles, so the area of the triangle is 1 2*r 2*(sin aob+sin boc+sin aoc).

The problem translates to finding the maximum value of sin aob + sin boc + sin aoc.

The sum of these three angles is 360 degrees.

Uh, this... Got preempted again)

Next is the what-if approach.

First, assuming that AOB is a fixed angle, the sum of these three sinusoidal values can become.

sin aob + sin aoc + sin boc are then combined with the differential product.

sin∠aob+2cos(1/2(∠aoc+∠boc))sin(1/2(∠aoc-∠boc))

Because AOC+ BOC is greater than 90 degrees, the value of this equation is non-negative.

So it should be. aoc- boc=0, i.e. aoc= boc, the value of the equation is maximum.

In the same way, such a conclusion can be drawn for the other two corners.

So when these three angles are equal, the value of sin aob + sin boc + sin aoc is maximum.

Thus the area of the circle within the regular triangle is the largest.

Radius = 2 3 of the height of the regular triangle [The center of the circle is the center of gravity of the triangle] The radius is r, and the area of the circle = r

Regular triangular side, 2 3H=R, H=3 2R

h=√3/2 a-- a=2√3/3h

s△=1/2ah=1/2 x 2√3/3x3/2r x 3/2r =3√3/4r²

Area ratio: 3 3 4:

A triangle must satisfy the other if it satisfies any of the following, and a triangle with equal three sides or equal triangles is an equilateral triangle:

1.The three sides are equal in length.

2.The number of three internal angles is 60 degrees.

3.An isosceles triangle with an internal angle of 60 degrees.

Because the area of the circumscribed triangle of the circle is equal to: s=1 2absinc=abc 4r

r is the radius of the circle), and the area is largest when a=b=c.

Let the radius of the circumscribed circle of the triangle ABC be r, then the triangle of s abc=(1 2)absinc=2r 2sinasinbsinc

2r 2[(sina + sinb + sinc) 3] 3 (mean inequality) 2r 2 3 = (3 3 4) r 2 (qinsheng inequality) When sina = sinb = sinc, i.e. a = b = c, the equal sign holds, so the area is maximum when the triangle is a regular triangle.

The circumscribed triangle of the circle, when an edge is determined within the circle, then when the perpendicular line of this side is made within the circle, only the line segment passing through the center of the circle is the longest.

A triangle is formed by connecting the intersection of the perpendicular line with the intersection of the known edges on the circle.

According to the area formula of the triangle s triangle = the base height of the triangle 2, so the known line segment is unchanged, and the larger the height, the larger the area of the triangle.

Radius = 2 3 of the height of the regular triangle [The center of the circle is the center of gravity of the triangle] The radius is r, and the area of the circle = r

Regular triangular side, 2 3H=R, H=3 2R

h=√3/2 a-- a=2√3/3h

s△=1/2ah=1/2 x 2√3/3x3/2r x 3/2r =3√3/4r²

Area ratio: 3 3 4:

A triangle must satisfy the other if it satisfies any of the following, and a triangle with equal three sides or equal triangles is an equilateral triangle:

1.The three sides are equal in length.

2.The number of three internal angles is 60 degrees.

3.An isosceles triangle with an internal angle of 60 degrees.

For any triangle within a circle, when one side is fixed, the area of the triangle on the same side of this side must be greater than the area of the other two sides if the other two sides are equal. That is, the fixed side is the base, and on the same side of the bottom edge, the area of the inner isosceles triangle is greater than that of the non-isosceles triangle.

After obtaining an isosceles triangle, a new isosceles triangle is constructed with a waist as the base, and the area of the new isosceles triangle will be a little larger. And so on, and if you continue to construct like this, you will get infinitely close to an equilateral triangle.

The rigorous proof process should be as follows: first, it is proved that for any non-isosceles triangle, an isosceles triangle can always be found to have a larger area than it; Secondly, it is proved that the area of any isosceles triangle must be smaller than that of the equilateral triangle. This.

Because it is an equilateral triangle, it is more special, the center of the circle is also the inner and outer center of the triangle, the distance to the three vertices on the three sides is equal, and it is a triangular bisector, the angles are divided into thirty degrees, and a perpendicular line is made through the center of the circle, and a triangle will appear, the hypotenuse is the radius, there is an angle of thirty degrees, and the side of the angle pair is a concentric line, which is, that is, the height of one-third (), the height is 1,5r, and the other right-angled side is 3r, so the side length of the inner triangle is 2 3r, and the area is.

Because that triangle is the largest at both the height and the bottom.