Geometry questions, first year of junior high school. Geometry questions in the first year of junior

By the inscription: a + c + 1 = 180°

a+∠d+∠2=180°

b+∠d+∠3=180°

b+∠e+∠4=180°

c+∠e+∠5=180°

So: 2 ( a + b + c + d + e) + 1 + 2 + 3 + 4 + 5 = 180° 5

a+ b+ c+ d+ e=180° 3So:2( a+ b+ c+ d+ e)=180° 2So: a+ b+ c+ d+ e=180°<>

It's not perfect, it's just a different shape.

Both graphs are stable.

If it is a separate 6-sided and 5-sided shape, there is no stability, but when one or several straight lines are added to the figure, the figure is limited and has stability, and the principle is the same as that of a triangle.

If you take two sides of a triangle, the non-common endpoints of the two edges are connected by a third edge.

The third edge cannot be retracted or bent.

The distance between the two endpoints is fixed.

The angles between these two sides are fixed.

These two sides are optional.

The triangle is fixed at all three corners, which in turn holds the triangle in place.

The triangle has stability.

If you take two adjacent edges of an n-sided (n 4), the non-common endpoints of the two edges are connected by more than one edge.

The distance between the two endpoints is not fixed.

The angle between these two sides is not fixed.

The n-sided (n 4) is not fixed at each corner, so the n-sided (n 4) has no stability.

In the same way, your polygon is actually a number of triangles with straight lines added.

So, they are stable.

I can't see the figure, I draw it according to the inscription, because the angle doa is y, so the angle dob is y, and the tangent angle boe is x, then the angle eoc is 2x, so 3x+2y=180 degrees.

x+y=70 degrees, so the system of continuous equations, then x=40 degrees, y=30 degrees; So the angle sought is 80 degrees.

Solution: Because be=cf

And because ec=ce

So BE+EC=CF+CE (property of the equation).

i.e. bc = ef

So in abc and def, ab=de

b=∠1bc=ef

So abc is equal to def(sas).

So ac=df (congruent triangles correspond to equal sides).

We're doing it little by little on the spot, I don't know if it's different from ours here, but that's how it was when we were in the first year of junior high school

be+ec=bc

cf+ec=ef

ef=bcab=de

Angle b = angle 1 so according to the corner edge theorem.

The triangle abc is all equal to the triangle def

So ac=df

Proof: Because be=cf

Because BE+CE=CF+CE

So bc=ef

So in abc and def, ab=de

b=∠1bc=ef

So abc is equal to def(sas).

So ac=df

Note: "Therefore" and "all equals" should be written in symbolic form, and it should be known that the equal sides plus the parts contained in common are still equal.

Because be=cf

So be+ec=cf+ec (the property of the equation) is because ab=de

b = 1, so the triangle abc is all equal to the triangle def (the corner edges), so ac = df (the corresponding sides of the congruent triangle are equal).

You can draw a diagram, the first time, the area is the original 1 2, the second time, the area is the original 1 4, the third time is 1 8, the fourth time is 1 16, the square of the circumference ratio = the area ratio, so it is 1 4

On the first floor, I haven't learned anything like that in the first year of junior high school!!

The first congruent triangle has been learned.,Similar triangles haven't been.。。。