The first day of junior high school, ask for help!! In the second year of junior high school, the ma

dab=180-30=150

cba=180-100=80

dcb=180-120=60

The sum of the inner angles of the quadrilateral is 360 degrees.

So d=360-150-80-60=70

It is known that the degrees of the angle DAB, the angle ABC, and the angle BCD are 150 degrees, 90 degrees, and 60 degrees, because the complementary angular sum is 180 degrees.

The sum of the inner angles of the quadrilateral is 360 degrees, so the angle d is 60 degrees.

Angle D + Angle C + Angle ABC + Angle BCD = 360

70° from the outer angle

yzwb I love my family to swipe the ticket the most.

Because the outer angles of angle DAB, angle ABC, and angle BCD are 30 degrees, 100 degrees, and 120 degrees (known) respectively

So the number of outer angles of angle D is 360 degrees, minus the outer angle of the angle abc, the outer angle of the ab, the outer angle of the subtracted angle of the ab, and the outer angle of the reduced angle of the bcd = 110 degrees.

So the degree of angle d is equal to 180 degrees minus 110 degrees = 70 degrees.

The sum of an angle and its outer angle is 180 degrees.

From this, the three angles of the quadrilateral ABCD can be found to be A=150, B=80, and C=60, respectively

From the sum of the inner angles of the quadrilateral to 360°, the degree of angle d = 360-150-80-60 = 70° can be obtained

The sum of the inner angles of the quadrilateral is 360, and according to the conditions, we know that we have told the three angles to find the fourth angle, which is to subtract the difference between the sum of the three angles from 360.

The outer angles are 360 in total, and after subtracting those three, there is 110, which is the outer angle of d, so d is 70 degrees.

Let the square of the high school bridge side length ab=bc=a

From the square nature, AC = AE = root Qi Meng, 2a, ac ef

It is derived from the sine theorem in ABE.

ab sin should not be a geometry problem in the second year of junior high school).

Let ab=1, because the square abcd, the front of the stove is called ab=bc=1, so ac=root number 2

Because of the diamond aefc, AC=AE=root number 2, and because AB=2, AE=root number 2

EBA = 90 degrees.

Therefore, according to the formula, divide the adjacent side by the hypotenuse and the sum of the beveled edge, and find the degree of jujube.

Because the ABCD square friend group closed AC is diagonal, so the angle CAB=45°, the angle CBA=90°, and because the four-good split-edge AEFC is or good diamond, so AC EF, that is, AC BF, so the angle ACB = angle FBC

When the point E is on the outside of a square ABCD, it is obtained from ABCD being a square and ADE being an equilateral triangle.

cde＝90°＋60°＝150°，de＝ad＝dc，∠dec＝∠ecd＝（180°－150°）÷2＝15°

The same can be done for AEB 15°

then bec aed aeb dec 60° 15° 15° 30°

a≤boc＜180

As long as o is in the triangle Lu Nai abc this plane class is such a hand circle. When O coincides with A, then Boc = A, and if O is infinitely (in) close to BC, then Boc is close to 180

The angle BOC degree is greater than the angle A, less than 180 degrees.

Knowing that the point ABC is any three points on the same line, AC=7cm, BC=3cm, then the distance between the midpoint line of the line segment AC and BC is ().

1.（ac+bc）/2=10/2=5cm

It is known that the line segment ab=6cm, draw the line segment bc=4cm on the straight line, if m, n are the midpoints of ab and bc respectively, 1Find the distance between m and n;

m is the midpoint of AC.

bm=1/2ab=3cm

In the same way, bn=1 2bc=2cm

mn=bn+bm=5 cm

Another possibility.

m is the midpoint of AC.

bm=1/2ab=3cm

In the same way, bn=1 2bc=2cm

mn=bm-bn=1 cm

2. If AB=ACM, BC=BCM, and the rest of the conditions remain unchanged, what is the distance between M and N (A B)?

m is the midpoint of AC.

bm=1/2ab=a/2cm

In the same way, bn=1 2bc=b 2cm

mn=（a+b）/2 cm

Another possibility.

m is the midpoint of AC.

bm=1/2ab=a/2cm

In the same way, bn=1 2bc=b 2cm

mn=（a-b）/2 cm

3. Analyze the solution process of (1) and (2), and what rules do you find from it?

The distance between the midpoints of the two line segments is equal to half of the sum of the line segments.

It should be this.

de is the perpendicular bisector of bc, be=cethen ae + be = ae + ce = ac = 8

So: ab = perimeter of abe - (ae + be) = 14 - 8 = 6