Find the degree of DCE and the degree of BAE

Because ad=ac, so cde= acd, so cde+ bcd=90 degrees (because acb=90 degrees);

Because be=bc, so bec= ecb, so bec+ ace=90 degrees (because acb=90 degrees);

So cde+ bcd+ bec+ ace=180 degrees.

Because DCE+ CDE+ BEC=180 degrees (the sum of the inner angles of the triangle is 180 degrees), DCE= BCD+ ACE

So DCE=45 degrees.

45 degrees Let a=x then b=90-x so ceb=(180-(90-x)) 2 adc=(180-x) 2 so edc=180- ceb- adc=45 degrees.

Because the angle ACB is a right angle, the angle A + the angle B = 90 degrees.

And because AC=AD, the angle ACD=AngleADC=180-AngleAA=BC, so the AngleBCE=AngleBEC=180-AngleBSo, 180-2 Angle, CEB+180-2, Angle, BEC=90, so the AngleCeb+AngleADC=135

So the angle DCE=180-135=45

There are two ways to do this:

1.Keep it simple.

ac=ce ∴∠cae=∠e

and cae+ e= bca=45° (the outer angle is equal to the sum of the two inner angles) cae= e=

BAC = 45° (triangle BAC is an isosceles right triangle with an acute angle of 45°) BAE = BAC + CAE = 4th floor The last step is written incorrectly) 2You can draw a picture to see.

Since it is a square ABCD, you may want to assume that the side length of all four sides is 1

Now extend BC to E so that CE AC, so CE=1, BE=2.

So in the triangle bae, ab=1, be=2, b=90°, so tg bae=be ab=2, bae=arctg2=hope it can help you.

tan(bae)=1:2 and inverse trigonometric functions can deduce the angle bae=degrees.

Degrees Reverse extension doesn't seem to be calculated Normal extension AC AE = 90 + 45 = 135 degrees when AE is in the same direction AC=AE CAE = degrees I made a mistake just now I'm sorry 45 degrees = degrees of 90 degrees.

Solution: It's abe=32°. Filial piety.

If yes, the answer is as follows.

CD EF DCE + CEF = 180° (two straight lines are parallel, complementary to the side inner angles) CEF = 180°-160° = 20°

ab cd abe= bef (two straight lines with equal headskin angles) bef= bec+ cef

abe= bef=20°+ bec=32° bec=12°

Because AD=AC, the crack is cDE= ACD, so CDE+ BCD=90 degrees (because ACB=90 degrees);

Because be=bc, so bec= ecb, so bec+ ace=90 degrees (because acb=90 degrees);

So cde+ bcd+ bec+ ace=180 degrees.

Because DCE+ CDE+ BEC=180 degrees (triangle.

Inner corners. and for 180 degrees of agitation after the end).

So Mingqin DCE= BCD+ ACE

So DCE=45 degrees.

Because the angle ACB is a right angle, the angle A + the angle B = 90 degrees.

And because of the fierceness of things.

AC=AD so angle ACD=Angle ADC=180-Angle ABE=BC so Angle BCE= Angle BEC=180-Angle B So, 180-2 Angle Cover Lingqiao CEB+180-2 Angle BEC=90 So. Angular ceb + angular adc = 135

So the angle DCE=180-135=45

2=180°-120°=60° Sell Jane.

Answer: The degree of 1 is 25°; The degree of fibrillation 2 is 60°

Triangle inner angle and 180 degrees. Several vertices can be connected regardless of the polygon. Make a few triangles, and you're done.

As you see in that figure, connect GK HK IK JK. It became the sum of the inner angles of the four triangles. That's 4 times 180 equals 720 degrees.

Hope...

n-1)*180 The inner angle of the polygon and the formula n are the number of sides, and when n is taken as 6, it is 720 degrees.

There's always a prerequisite!

This question is left to me, in order to avoid me doing it for a long time, but you give it to others, so I will help you answer it now, you give it now, you don't know how to disturb the question, but it is best to add a lot of points.

According to the theorem: the outer angles are equal to the sum of two non-adjacent inner angles.

2 = 90 degrees + 65 degrees = 155 degrees.