How to draw an angle equal to a known angle with a ruler and a compass

How to draw an angle equal to a known angle with a ruler and a compass

The method of drawing an angle equal to the known angle according to the ruler is as follows:
1. Any ray o'x
2. Take the vertex o of the known angle as the center of the circle and any length as the radius, draw an arc to intersect the two sides of the known angle at two points a and B respectively
3. Draw an arc with o 'as the center and OA length as the radius, and intersect the ray o'x at point a'
4. Draw an arc with a 'as the center and ab length as the radius, intersect with the arc drawn in step 3 at point B', and connect with o'b ', then ∠ a'o'b' is the angle
From the above steps, we can see that o'a '= OA, o'b' = ob, a'B '= AB, so △ a'o'b' ≌ △ AOB, that is ∠ a'o'b '= ∠ AOB. Here we prove the congruence of triangles by using the edge theorem

What is the basis for the angle drawn with ruler and gauge to be equal to the known angle
What is the basis for making an angle equal to a known angle with a compass and a ruler?
Students say it's "SSS"

It's "side by side."
It should be written in the book

It is known that a (4, m) B (- 1, n) is on the image with inverse scale function y = 8 / X
What is the function expression of line AB?

Put two points in
So m = 8 / 4 = 2
n=8/(-1)=-8
A(4,2),B(-1,-8)
Let AB be y = KX + B
SO 2 = 4K + B
-8=-k+b
subtract
5k=10
k=2
b=2-4k=-6
So y = 2x-6

It is known that the image of the inverse scale function y = m − 8x (M is a constant) passes through point a (- 1,6). (1) find the value of M; (2) as shown in the figure, the line AC passing through point a intersects the image of the function y = m − 8x at point B, intersects the X axis at point C, and ab = 2BC, and finds the coordinates of point C

(1) The solution of ∵ image passing through a (- 1,6) and ∵ m − 8 − 1 = 6 is m = 2, so the value of M is 2; (2) the perpendiculars of x-axis passing through a and B respectively are e and D. according to the title, AE = 6, OE = 1, namely a (- 1,6), ∵ BD ⊥ x-axis, AE ⊥ x-axis, ∥ AE ∥ BD, ∥ CBD ∥ CAE, ∥ CBCA = BD

As shown in the figure, the image of the first-order function y = KX + B intersects with the image of the inverse scale function y = x / m at two points a (- 2,1), B (1, n)
1. Find the analytic expressions of inverse proportion function and linear function;
Write the value range of X that makes the value of the first function greater than that of the inverse function according to the image
Netizens, do me a favor. I can't draw pictures,

A (- 2,1) on the inverse scale function, that is, M = - 2 B (1, n) on the inverse scale function, n = - 2 B (1. - 2)
AB is introduced to solve the system of equations on a linear function and K = - 1, B = - 1, y = - X-1
The equation - X-1 ＞ - 2 / x 2 / x-x-1 > 0 can be solved without drawing
When x > 0, x ^ 2 + X-2

As shown in the figure, the graph of the first-order function y = KX + B and the inverse scale function y = m / X intersects a (2,3), B (- 3, n)
1. Find the relationship between the first order function and the inverse proportion function;
2. According to the given conditions, please write the solution set of inequality KX + b > m / X directly______ ；
3. Take BC ⊥ X axis through point B, take point C as the vertical point, and find s △ ABC

The graph of the first-order function y = KX + B and the inverse scale function y = m / X intersects at two points a (2,3), B (- 3, n). The graph of the first-order function y = KX + B and the inverse scale function y = m / X intersects at two points a (2,3), B (- 3, n). The graph of the first-order function y = KX + B and the inverse scale function y = m / X intersects at two points a (2,3), B (- 3, n). The graph of the first-order function y = KX + B and the inverse scale function y = m / X intersects at two points a (2,3, n)

There is a point P (m, n) on the image of inverse scale function y = K / x, M + n = 3
And the distance of the far point of p-channel is root 13, then the expression of inverse scale function ()

According to the distance from P to the origin as root 13, we can get
m^2+n^2=13
And M + n = 3
The solution of the equations is obtained
m=[3-（√17）]/2 n=[3+（√17）]/2
Or M = [3 + (√ 17)] / 2, n = [3 - (√ 17)] / 2
It can be obtained by substituting it into the analytic expression of the function
k=mn=-2
Then the function expression is
k=-2/x

It is known that the two points of M. n are symmetric about the Y axis, and the point m is on the image of inverse scale function y = k divided by X. the coordinates of the point m are set as a, B on the image of primary function y = x + 3
Then the quadratic function y = ABX ^ 2 + (a + b) x -------?

M(a,b)
M. If n points are symmetric about y axis, then n (- A, b)
Point n is b = - A + 3 on the image of linear function y = x + 3
Point m b = K / A on the inverse scale function y = K / X
ab=k,a+b=3
Substituting quadratic function y = ABX ^ 2 + (a + b) x
We get y = KX ^ 2 + 3x

It is known that the images of inverse scale function y = K / X and linear function y = ax + B pass through point P (2, - 1), and when x = 1, the two functions
It is known that the images of the inverse scale function y = K / X and the first-order function y = ax + B pass through P (2, - 1), and when x = 1, the values of the two functions are negative reciprocal. The analytic expressions of the two functions are obtained

x=2,y=-1
So - 1 = K / 2
-1=2a+b
Then k = - 2
b=-1-2a
x=1
y=k/x=-2
The values of the two functions are negative reciprocal
So x = 1
y=ax+b=1/2
That is, a + B = 1 / 2
a+(-1-2a)=1/2
a=-3/2
b=2
So y = - 2 / X
y=-3x/2+2

It is known that an intersection point of the image with inverse scale function y = x / K and the image with primary function y = 2x + 4 is p (- 1, a) to find the relation of inverse scale function

A=-2+4=2
The coordinates of point P (- 1,2)
It can be obtained by substituting it into the inverse proportion function
k=-2
The relation of inverse scale function y = - 2 / X