As shown in the figure, the triangle ABC is inscribed with the circle O, ad bisects ∠ BAC, intersects ⊙ O and D, passes through D as de ∥ BC, and intersects the extension line of AC with E The triangle ABC is inscribed with the circle O, ad bisects ⊙ BAC, intersects ⊙ O and D, passes D as de ∥ BC, and intersects the extension line of AC with E Try to judge the position relationship between de and circle O and prove your conclusion 2. If ∠ e = 60 degrees, the radius of circle O is 4, and find the length of ab

As shown in the figure, the triangle ABC is inscribed with the circle O, ad bisects ∠ BAC, intersects ⊙ O and D, passes through D as de ∥ BC, and intersects the extension line of AC with E The triangle ABC is inscribed with the circle O, ad bisects ⊙ BAC, intersects ⊙ O and D, passes D as de ∥ BC, and intersects the extension line of AC with E Try to judge the position relationship between de and circle O and prove your conclusion 2. If ∠ e = 60 degrees, the radius of circle O is 4, and find the length of ab

1. De is tangent to circle o
Because ad bisects ∠ BAC, so ∠ bad = ∠ CAD, so arc BD = arc CD, connecting do, then do vertically bisects BC, because de / / BC, so od is vertical De, so De is tangent to circle o
2. Because de ∥ BC, so ∠ BCA = ∠ e = 60 degrees, so ∠ AGB = 60 degrees, and because Bo is diameter, so ∠ bag = 90 degrees, so AB = (8 square - 4 square) arithmetic square root, so AB = 4 roots 3

As shown in the figure, the diameter ab of ⊙ o is 10cm, the chord AC is 6cm, the bisector of ⊙ ACB intersects ⊙ o in D, and find the length of BC, ad, BD

 AB is the diameter  ACB = ∠ ADB = 90 ° in RT △ ABC, AB2 = ac2 + BC2, ab = 10cm, AC = 6cm ᙽ BC2 = ab2-ac2 = 102-62 = 64 ᙽ BC = 64 = 8 (CM) and CD bisection

As shown in the figure, triangle ABC is inscribed in circle O, ad bisector angle BAC intersects circle O in D, De is made through D, parallel to BC, and extension line crossing AC is at E 1. Try to judge whether De is tangent to circle o 2. If the angle e = 60 degrees and the radius of circle O is 4, find the length of ab

1. Connect OD, BD and CD
∵∠BAD=∠CAD
﹤ BD = CD (equal circular angles to equal chords)
And ob = OD
⊥ BC (vertical line)
And de ‖ BC
∴DE⊥OD
⊙ De is tangent to ⊙ o
2.∵∠ACB=∠E=60°
The length of the corresponding chord AB is constant
Assume that AC is the diameter (where the length of AB is most easily calculated)
Then △ ABC is a special right triangle
AB=4√3

Given the chord AB = AC of circle O, the bisector BD of ∠ ABC intersects the circle O at point D, the extension line of AD and BC intersects at point E, ∠ BAC = 50 ° to find the degree of ∠ E

Let me have a try
AB=AC,∠BAC=50°
From the sum of the inner angle and 180, we can get ∠ ACB = ∠ ABC = 65
The bisector BD of ∠ ABC obtains ∠ DBD = 1 / 2 ∠ ABC = 32.5
∵ DAC = ∵ DBC = 32.5 (circular angle opposite to the same arc)
∴∠E=∠ACB-∠DAC=32.5

As shown in the figure, point D is on the extension line of diameter ab of ⊙ o, point C is on ⊙ o, and AC = CD, ﹤ ACD = 120 ° and CD is the tangent line of ⊙ o: if the radius of ⊙ o is 2, the area of shadow part in the figure is______ .

Connected to OC, ∵ AC = CD,  ACD = 120 °,  CAD = ∵ d = 30 °,

AB is the diameter of circle O, CA tangent circle O to a, CB intersect circle O to D, if CD = 2, BD = 6, then SINB value Because I just got into touch with trigonometric function, I don't understand it

Connecting ad
∠CAD=∠B
∠CDA=∠CAB=90°
△ACD∽△CAB
AC：BC=CD：AC
AC²=CD×BC=2×8
AC=4
sinB=AC/BC=4/8=1/2

As shown in the figure, AB is the diameter of ⊙ o, AC is the chord, CD is the tangent of ⊙ o, C is the tangent point, and ad ⊥ CD is at point D （1）∠AOC=2∠ACD； （2）AC2=AB•AD．

It is proved that: (1) ∵ CD is tangent of ⊙ o,  OCD = 90 °,
That is ∠ ACD + ∠ ACO = 90 °. ① (2 points)
∵OC=OA，∴∠ACO=∠CAO，
Ψ AOC = 180 ° - 2 ∠ ACO, that is ∠ AOC + 2 ∠ ACO = 180 °,
Divide both sides by 2 to get: 1
2 ∠ AOC + ∠ ACO = 90 °. ② (4 points)
From ①, ②, we get: ∠ acd-1
(5 points)
(2) As shown in the figure, connect BC
∵ AB is the diameter,  ACB = 90 °. (6 points)
In RT △ ACD and RT △ ABC,
∵∠AOC=2∠B，
∴∠B=∠ACD，
/ / RT ∽ ACD ᦻ RT △ ABC, (8 points)
∴AC
AB＝AD
AC, i.e. ac2 = ab · ad. (9 points)

It is known that: as shown in the figure, AB is the tangent line of ⊙ o, the tangent point is a, OB intersects ⊙ o at C and C is the midpoint of ob, and the chord CD passing through C makes

Connect OA, OD
∵∠DCA=45°
∴∠AOD=90°
Qi
The length of ad is 90 π· OA
180＝
Two
2 pi
∴OA=OD=
Two
∴AD=
OA2+OD2＝
4=2
∵ AB is ⊙ o tangent
∴OA⊥AB
C is the mid point of RT △ AOB beveled edge
∴AC=OC=OA=
2．

As shown in the figure, AB is the diameter of ⊙ o, AC is the chord, CD is the tangent of ⊙ o, C is the tangent point, and ad ⊥ CD is at point D （1）∠AOC=2∠ACD； （2）AC2=AB•AD．

It is proved that: (1) ∵ CD is tangent of ⊙ o,  OCD = 90 ° i.e. ? ACD + ACO = 90 °. ① (2 points) ? OC = OA, ∠ ACO = ∠ Cao, ∠ AOC = 180 ° - 2 ∠ ACO, that is ? AOC + 2 ∠ ACO = 180 ° divided by two sides, 12 ﹤ AOC + ACO = 90 ° is obtained

As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, CE is the diameter of ⊥ o, CD ⊥ AB and D are perpendicular feet. It is proved that ⊙ ACD = ∠ BCE

Proof: connect EB,
∵CD⊥AB，
∴∠ADC=90°，
∴∠A+∠ACD=90°，
∵ CE is the diameter of ⊙ o,
∴∠CBE=90°，
∴∠E+∠ECB=90°，
∵∠A=∠E，
∴∠ACD=∠BCE．