It is known that AB CD, BE bisected ABC, DE bisected ADC verification 5

Certificate]: Description: Because you can't draw a picture, please draw the figure yourself.

Let bc and de intersect at the point o. Note boe= 5, so cod= boe= 5 [equal to the vertex angles].

From the meaning of the title, it can be seen that ebo=1 2 b, cde=1 2 d

And because :ab cd, there is:

a=∠d,∠c=∠b.[The two straight lines are parallel, and the inner wrong angles are equal].

So ebo=1 2 b=1 2 c,, cde=1 2 d=1 2 a

In BEO, e=180°-(5+ ebo)【*

In COD, 5=180°-(C+ Cde).

Replace ebo=1 2 c, cde=1 2 a, 5=180°-(c+ cde) into the formula [*] to simplify: e= .Certification]|

Commentary: The key to this problem is to find a bridge to connect e with other known angles, and eventually transition to a and c. Whether you can find the pair of cod= boe= 5 pairs of vertex angles and two triangles beo and cod is the key to the problem, and this is the breakthrough.

It can't be true, it should be wrong (if it's a square, then e 180°,

Brother fooled you, who is this question?

It's a false proposition that can't be proven at all.

Suppose that the square angle e=180° and the angles a and c are 90° each, then e=a+c ) and therefore e= is not true.

According to the polygon inner angle and the formula 180° (n-2).

So the sum of the internal angles of the quadrilateral ABCD and the quadrilateral BCDE is 360°

So abc+ cda=360-80-n=280°-n°

So ebc+ cde=140°-n° 2

So bed+ bend bend=360-(140-N2)=220°+N°2

bed=220°-n° 2,2,180°(n-2)so the inner angles of the quadrilateral abcd and the quadrilateral bcde hail before and are 360°, so abc+ cda=360-80-n=280°-n°, so ebc+ cde=140°-n° 2, so bed+ bcd=360-(140-n 2)=220°+n° 2 bed=220°-n° 2,3,(2)n 2°+40°,2,ghnnfghn, 2, according to the polygon inner angle and the formula 180° (n-2).

So the sum of the internal angles of the quadrilateral ABCD and the quadrilateral BCDE is 360°

So abc+ cda=360-80-n=280°-n°

So ebc+ cde=140°-n° 2

So bed+ bcd=360-(140-n2)=220°+n°2

bed=220°-n°/2

It's really hard, 2, take a picture or draw one with a computer drawing board, and then send a picture, 1, as shown in the figure, know ab parallel cd, be bisected abc, de bisect adc, bad=80°, try to find:

1.I'll do it, I don't need to.

2.If bcd=n°, try to find the degree of the bed.

The picture is on the test paper, and the hair is not buried. Solving.

Witness the extension of the CE to BA at point F

CE bisected round silver BCD

dce＝∠bce

ab cd f muffled dce, fad dbc bfbe bisected abc

EF CE (3-in-1).

aef≌△dec （aas）

af＝cdbf＝ab+af＝ab+cd

bc＝ab+cd

The Math Tutoring Team answered your questions

According to the polygon inner angle sum formula 180°(n-2), so the sum of the internal angles of the quadrilateral ABCD and the quadrilateral bcde is 360°, so abc+ cda=360-80-n=280°-n°, so ebc+ cde=140°-n° 2, so bed+ bcd=360-(140-n2)=220°+n° 2

bed=220°-n°/2

180°(n-2) so the sum of the internal angles of the quadrilateral ABCD and the quadrilateral bcde is 360°, so abc+ cda=360-80-n=280°-n°, so ebc+ cde=140°-n° 2, so bed+ bcd=360-(140-n2)=220°+n° 2, bed=220°-n° 2

Take a picture or draw one on a computer tablet and send a picture.

The answer to the first question is simple, it's 50.

The second question is that the degree of the angle bed is x.

It is easy to know that bed= ebc+ bcd+ cde, then ebc+ cde=x-n, and be bisected angle abc, de bisected angle adc, there is abc+ adc=2( ebc+ cde)=2(x-n), and dab+abc+bcd+cda=360, that is, 80+2(x-n)+n=360, x=140+n 2.

Let's look at the quadrilateral abed first, and the four angles add up to 360°

So bed=360°- bad- abe- adeabe=1 2 abc ade=1 2 adc, so bed=360°-80°-1 2 abc-1 2 adc=280°-1 2( abc + adc).

When looking at the quadrilateral abcd, the four angles add up to 360°, so abc+ adc=360°- bad- bcd=360°-80°-n°=280°-n°

So bed=280°-1 2(280°-n°)=140°+1 2 n°

The general situation has been written clearly, and the specific conclusion still needs to be organized by yourself.

Certificate: ab cd dcb abc 180° and ce is dcb angle bisector be abc angle bisector 1 2 90° certification.

ab cd (known), abc + bcd = 180° (two straight lines are parallel, complementary to the side inner angles) and be bisected abc, ce bisected bcd, (known) 1+ 2 = 1 2 abc + 1 2 bcd = 1 2 180° = 90° (bisector definition). At the end of the service, please rate it!