As shown in the figure, the four points a, B, C, D are on the circle O. if an external angle DCE of the quadrilateral ABCD is 70 °, calculate the size of the angle bad and the angle BOD

As shown in the figure, the four points a, B, C, D are on the circle O. if an external angle DCE of the quadrilateral ABCD is 70 °, calculate the size of the angle bad and the angle BOD

: (the outer angle of the inscribed quadrilateral of a circle is equal to the inner diagonal)
∵ an outer angle of the quadrilateral ABCD  DCE = 70 °,
∴∠A=∠DCE=70°,
∴∠BOD=2∠A=140°

As shown in the figure. A.b.c.d. is a point on the circle O and ∠ bcd100 ° is used to calculate the size of ∠ BOD and ∠ bad

The angle BOD is the center angle corresponding to the angle BCD
therefore
Angle BOD = 2 angle BCD = 200
But define the angle as the angle less than 180 degrees
therefore
Angle BOD = 160
Angle bad = 100
This is the same as the circumference angle of arc BD

As shown in the figure, ABC is three points on the circle O and ∠ AOE = 90 °, then the degree of ∠ ACB is

What is e? Where is it?

As shown in the figure, ⊙ o three points a, B, C, ab = AC, bisector of ⊙ ABC intersects ⊙ o at point E, ⊙ o intersects ⊙ o at point F, be and CF intersects at point D. is the quadrilateral afde diamond? Verify your conclusion

Conclusion: the quadrilateral afde is rhombus. It is proved that  ABC = ∠ ACB,  Abe = ∠ EBC = ∠ ACF = ∠ FCB. And ∠ Fab and ∠ FCB are the circular angles on the same arc, ? Fab = ∠ FCB, the same reason ∠ EAC = ∠ EBC

As shown in the figure, triangle ABC is inscribed in circle O, AB is the diameter, bisector of ∠ CBA intersects AC at point F, circle O at point D, de ⊥ AB at point E and AC at point P If the radius of the circle is 5, AF = 15 / 2, find Tan ∠ ABF

With AD
∠CAD=∠CBD=∠ABD
∠ADB=90
So there are
Triangle abd is similar to triangle AFD
AB/AF=AD/DF=10/7.5 = 4/3
tan∠ABF = tan∠FAD = 3/4

As shown in the figure, on the circle O, three points a, B, C, ab = AC, the bisector of ∠ ABC intersects the circle O and the point E, the bisector of ∠ ACB intersects the circle O with the point F, be and cf. thank you

It is proved that because AB = AC, the angle ABC = ACB, because CF and be are bisectors, so the angle ACF = FCB = EBA = angle EBC, so the angle AFC + angle FAE = angle ABC + FCB + FCA + BAC = 180, so AE is parallel to FC, similarly FA is parallel to EB, so FDEA is parallelogram, and because angle EBA = FCA, FA = EA, so parallelogram FDEA is diamond

As shown in the figure, △ ABC is the inscribed triangle of circle O, and the bisectors of ab ≠ AC, ∠ ABC and ∠ ACB intersect the circle O at points D and e respectively, and BD = CE, then ∠ A is equal to () A. 90° B. 60° C. 45° D. 30°

Connect ad, be, ∵ BD = CE  arc BD = arc CE, ? BD = CE  arc BD = arc CE, ? bad =  cab, ? Abe, ? CBD, ? CAD = ∠ CBD (in the same circle, the circle angles opposite to the same arc are the same), cab = ∠ abd + ∠ Abe, ? AB

As shown in the figure, in the RT triangle ABC, angle c = 90, the bisector of angle ABC intersects AC at point D, O is a point on AB, and circle O passes through two points B and D, It is proved that AC is the tangent of circle o

Because it's a circle
So ob = od = radius
So angle ODB = angle OBD (isosceles)
So angle OBD = angle DBC = angle ODB
So OD ‖ BC
And the angle c is 90 degrees, so OD ⊥ AC
That is, AC is the tangent of the circle

As shown in the figure, ∠ A is the circumference angle of ⊙ O and ∠ a = 40 ° to find the degree of  OBC

∵∠A=1
2∠BOC，
And ∵ a = 40 °,
∴∠BOC=2∠A=80°，
In isosceles △ OBC, ∠ OBC = 180 ° - ∠ BOC
2=180°-80°
2=50°．

It is known that, as shown in the figure, a, B and C are three points on ⊙ o, ∠ oba = 50 ° and ∠ OBC = 60 °, then ∠ OAC=______ Degree

∵OA=OB，
∴∠OBA=∠OAB=50°，∠O=180°-2∠OBC=80°，
∴∠C=1
2∠O=40°，
∴∠CAB=180°-∠C-∠OBC-∠OBA=180°-40°-60°-50°=30°．
∴∠OAC=∠OAB-∠CAB=20°．