If the square ABCD is known, and the CD is used as the side for the equilateral CDE, then the degree

Solution: There are two scenarios:

When E is in a square ABCD, square ABCD, AD=CD, ADC=90°, equilateral CDE, CD=DE, CDE=60°, ADE=90°-60°=30°,AD=DE, DAE= AED= 1 2(180°- ADE)=75°;

When E is outside the square ABCD, the equilateral triangle cde, edc=60°, ade=90°+60°=150°, aed= dae= 1 2(180°- ade)=15°

So the answer is: 15° or 75°

Thank you.

If the equilateral cde is inside the square ABCD, then AED=(180°-30°)2=75°

If the equilateral cde is outside the square ABCD, then AED=30° 2=15°

When e is in a square ABCD, the square ABCD, AD=CD, ADC=90°, and equilateral CDE,CD=DE, CDE=60°, ADE=90°-60°=30°,AD=DE, DAE= AED= 1 2(180°- ADE)=75°;

When E is outside the square ABCD, the equilateral triangle cde, edc=60°, ade=90°+60°=150°, aed= dae= 1 2(180°- ade)=15°

AED = 15° or 75°

Draw your own ** to decide maybe 15 or 75

There are two cases of caution: (1) when E is in a square ABCD, as shown in Figure 1 square ABCD, AD=CD, ADC=90°, equilateral CDE, CD=DE, CDE=60°, ADE=90°-60°=30°, AD=DE, DAE= wide Lee AED=12(180°- ADE)=75°; (2) When E is in the positive opening of the square A....

The quadrilateral ABCD is the square shape of the noisy square, and the CDE is the equilateral missing meniscus triangle, AD=CD=DE; ∠ade=90°+60°=150°．

aed=（180°-150°）÷2=15°．

So the answer is 15°

ABCD is a square.

CDE is an equilateral triangle quietly destroyed.

ad=de=cd

adc=90°

cde=60°

ADE= Sakura-pure ADC+ CDE=90°+60°=150°AED= AED

AED = (180°- Ridge Movement Ade) 2=(180°-150°) 2=15°

b=3 a, because the parallelogram is equal to the diagonal 4 horns, and the sum of the 4 angles = 360

So c= a

d=∠b∠a+∠b+∠c+∠d）=360

c+3∠c+∠c+3∠c=360

8∠c=360

c = 45 degrees.

If it helps you, please remember, o( o thank you.

75° if E is within a square ABCD;

Solution: The triangle CDE is an equilateral triangle.

The three angles are all 60 degrees, and when AE is connected, the sum of the internal angles of the quadrilateral AECD is (n-2)*180=(4-2)*180=360 degrees.

15° if e is outside the square

Solution: AED inner angle and 180°, angle ADE=60+90=150°(180-150)2=15°

Draw a picture yourself and you will understand, if you don't understand, ask again! Please adopt it in time.

There are two scenarios:

Point E is inside the square ABCD: AED = 75°

Point E is outside the square ABCD: AED = 15°

When e is external, aed=(180-150)2=15

When e is internal, aed=(180-30) 2=75

60° angle BCE = 60 + 90 = 150°

Angle cbe = angle ceb = 15°

The triangle BCM is equal to the triangle DCM

Angular CDM = Angular CBE = 15°

Angle ADM = 90-15 = 75°, angle MAD = 45° angle AMD = 60° to select B

Solution: 60 degrees.

The specific solution is to see the ** file, left-click to enlarge, and it is not clear that you can continue to copy it into word with right-clicking.

<> this question belongs to the problem of good classification and discussion of Xiangque:

adc=90°, edc=60°, ade=30°, de=dc=da, aed= ead=(180°- ade) 2=75°

ADC=90°, EDC=60°, ADE=150°, and DE=DC=DA, AED= EAD=(180°- ADE) 2=15°

In summary, AED = 75° or 15°

ABCD is a square.

CDE is an equilateral triangle.

ad=de=cd

adc=90°

cde=60°

Guess Duan Zhaobu ade= adc + cde = 90° + 60° = 150° ead = aed

Grip aed = (180°- ade) 2 = (180°-150°) 2 = 15°

Flowers fall, rain and mud are cherished, and the well is borrowed from the work or into the investigation is still a good time to dust.

There are two scenarios:

square abcd,ad=cd, adc=90°,equilateral cde,cd=de, cde=60°, ade=90°-60°=30°,ad=de, dae= aed=1

180°-∠ade）=75°；

Equilateral triangle cde, edc=60°, ade=90°+60°=150°, aed= dae=1

180°-∠ade）=15°．

So the answer is: 15° or 75°