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As shown in the figure, take the diameter BC of ⊙ o as an equilateral ⊙ ABC, AB and AC intersect ⊙ o at D and e. it is proved that BD = de = EC
Proof: connect od and OE as shown in the figure
∵ △ ABC is an equilateral triangle,
∴∠B=60°．
And ∵ ob = OD,
The OBD is an equilateral triangle,
∴∠BOD=60°．
Similarly, △ EOC is an equilateral triangle, then ∠ EOC = 60 °
∵ BC is the diameter of ⊙ o,
∴∠DOE=180°-∠BOD-∠EOC=60°，
Qi
BD=
DE=
EC，
∴BD=DE=EC．
A C is an equilateral triangle, BC is the diameter of ⊙ O. the AB and AC edges intersect ⊙ o at D and e respectively. It is proved that the Arc de = arc BD = arc EC
Connect OD, OE; because ABC is an equilateral triangle; so angle ABC = angle ACB = 60; and ob = ob = OE = OC; so OBD, OCE is equilateral triangle; so angle DOB = 60, angle COE = 60; so angle DOE = 180 angle DOB angle COE = 180-60-60 = 60; so DOE is equilateral triangle; so angle BOD = angle DOE = angle EOC; so Arc de = arc BD = arc EC
In the graph, △ ABC is divided into four small triangles, of which the areas of three triangles are 6cm 2, 12 cm 2 and 8 cm 2 respectively
Let the area of the shadow part be X,
8：x=12：6，
12x=48，
x=4，
Answer: the area of shadow part is 4 square centimeter
In the graph, △ ABC is divided into four small triangles, of which the areas of three triangles are 6cm 2, 12 cm 2 and 8 cm 2 respectively
Let the area of the shadow part be X,
8：x=12：6，
12x=48，
x=4，
Answer: the area of shadow part is 4 square centimeter
The area of the square in the figure is 20 square centimeters. The area of the smallest circle outside the square is______ .
Let the radius of the circle be r,
Then the area of the square is 2r2 = 20 square centimeters,
r2=10，
So 3.14 × 10 = 31.4 (square centimeter);
A: the area of the smallest circle outside the square is 31.4 square centimeters
So the answer is: 31.4 square centimeter
As shown in the figure, in △ ABC, ad intersects BC at point D, ∠ bad = 15 °, ADC = 4 ∠ bad, DC = 2bd (1) Find the degree of ∠ B; (2) . B = ∠ CAD
（1）∵∠BAD=15°，∠ADC=4∠BAD，
∴∠ADC=60°，
∴∠B=60°-15°=45°；
(2) It is proved that CE ⊥ ad is made by C and connected with EB
∵∠ECD=90°-60°=30°
∴DC=2ED，
∵DC=2BD，
∴ED=BD，
∴∠DBE=∠DEB=∠ECD=30°，
∴∠EBA=45°-30°=15°=∠BAD，
∴AE=EC=EB，
∴∠CAD=∠ABD=45°．
As shown in the figure, in △ ABC, ad intersects BC at point D, ∠ bad = 15 °, ADC = 4 ∠ bad, DC = 2bd (1) Find the degree of ∠ B; (2) . B = ∠ CAD
(1) (2) it is proved that through C, CE ⊥ ad to e, connecting EB ? ECD = 90 ° - 60 °; DC = 2ed, ∵ DC = 2bd, ? ed = BD, DBE = ∠ DEB = ∠ ECD = 30 °, EBA = 45 ° - 30 ° = 1
As shown in the figure, in triangle ABC, the angle ABC = angle BAC, the angle bisector of angle BAC intersects the extension line of BC at point D, if the angle ADC = 2 / 1, angle CAD, calculate the degree of angle ABC
The intersection of the angle BAC and BC is on the BC line. It is not on the extension line
In the triangle ABC, the angle c = 90 degrees, the angle B = 40 degrees, D is a point on BC, and the angle ADC = 60 degrees. Find the angle CAD
Angle ADC equals angle B + angle DAC
So the angle DAC is 20 degrees
Because the angle cab is 50 degrees
So the angle CAD is 30 degrees
As shown in Figure 9, in the triangle ABC, AC = BC, the extension of the diplomatic bisector of ∠ BAC intersects with the extension line of BC at point D. if ∠ CAD = 2 ∠ ADC, calculate the degree of ∠ ABC
Angle ABC = angle cab, angle ABC = angle DAB + angle ADC = (180 degrees angle cab) / 2 + angle ADC = (180 degrees angle ABC) / 2 + angle ADC, angle ADC = 90 degrees - 3 angles ABC / 2 (1) Angle CAD = angle cab + angle DAB = angle ABC + angle DAB = angle ABC + (180 degree angle ABC) / 2 = 90 degree + angle ABC / 2,90 degree + angle ABC / 2 = 2 angle ADC
RELATED INFORMATIONS
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- 15. As shown in Figure 1, in triangle ABC, OB and OC are bisectors of angle ABC and angle ACB respectively. If angle a = x degrees, what is the degree of angle BOC
- 16. As shown in the figure, BC is the chord of circle O, OD ⊥ BC is at point E, intersection arc BC is at point D, and point a is a moving point on the major arc BMC (not coincident with B and C). BC = 4, radical 3, ed = 2 are known (1) Find the radius of circle O (2) Find the degree of ∠ a (3) Find the position of point a when the shadow part takes the maximum value I can't figure it out
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