As shown in the figure, it is known that the triangle ABC is inscribed in the circle O, the point D is on the extension line of OC, SINB = 1 / 2, angle d = 30 ° (1) prove that ad is tangent. (2) if AC = 6, find the length of AD

As shown in the figure, it is known that the triangle ABC is inscribed in the circle O, the point D is on the extension line of OC, SINB = 1 / 2, angle d = 30 ° (1) prove that ad is tangent. (2) if AC = 6, find the length of AD

(1) According to the graph, B is an acute angle, SINB = 1 / 2, so B = 30 degrees, angle CBA and angle DOA correspond to the same arc, so angle DOA = 60 degrees, angle d = 30 degrees, so angle Dao = 90 degrees, that is, Da is perpendicular to OA, so ad is tangent to circle O. from the proof of (1), angle DOA = 60 degrees, so angle AOC is equilateral triangle, so AC = OA = 6,

As shown in the figure, it is known that △ ABC is inscribed on the circle O, D is on the OC extension line, and the angle B = 30 ° (1) verify that ad is the tangent line of circle O. (2) if AC = 6, find the length of AD

This question lacks one condition: CD = OC
prove:
Connect Ao, AC
Then Ao = co = radius
∠ AOC = 2 ∠ B = 60 ° [the center angle of the circle opposite to the same arc is equal to twice the circumference angle]
⊿ AOC is an equilateral triangle
∴OA=OC=AC,∠ACO=60º
∵OC=CD
∴AC=CD
∴∠CAD=∠D=½∠ACO=30º
∴∠OAD=∠OAC+∠CAD=90º
/ / AD is the tangent of circle o
∵OD=OC+CD=12,AO=6
∴AD=√（OD²-OA²）=6√3

It is known that in △ ABC, ∠ ACB = 90 ° AC = BC, the straight line Mn passes through point C, and ad ⊥ Mn is in D, be ⊥ Mn is in E Results: ① △ ADC ≌ △ CEB; ② de = ad-be

It is proved that: ① ∵ ACB = 90 °, be ⊥ CE, ad ⊥ CE,
∴∠BEC=∠ACB=∠ADC=90°，
∴∠ACE+∠BCE=90°，∠BCE+∠CBE=90°，
∴∠ACD=∠CBE，
In △ ADC and △ CEB
∠ADC＝∠BEC
∠ACD＝∠CBE
AC＝BC ，
∴△ADC≌△CEB（AAS）．
②∵△ADC≌△CEB，
∴AD=CE，BE=CD，
∴CE-CD=AD-BE，
∵DE=CE-CD，
∴DE=AD-BE．

As shown in the figure, it is known that ⊙ o is the circumscribed circle of ⊙ ABC, and ab is the diameter. If PA ⊥ AB, Po passes through the midpoint m of AC, it is proved that PC is the tangent of ⊙ o

Proof: connect OC,
∵PA⊥AB，
﹤ PA0 = 90 °. (1 point)
∵ Po passes through the midpoint of AC, OA = OC,
Ψ Po bisection ∠ AOC
﹤ AOP = ∠ cop. (3 points)
In △ Pao and △ PCO, OA = OC, ∠ AOP = ∠ cop, Po = Po
≌△ PCO. (6 points)
∴∠PCO=∠PA0=90°．
That is, PC is the tangent line of ⊙ O. (7 points)

It is known that: as shown in the figure, in △ ABC, D is a point on the edge of AB, and the circle O passes through the three points of D, B and C, ∠ doc = 2 ∠ ACD = 90 ° (1) Prove that O is a straight line; (2) If ∠ ACB = 75 ° and the radius of circle O is 2, find the length of BD

(1) It is proved that: ∵ od = OC, ∵ doc = 90 °,
∴∠ODC=∠OCD=45°．
∵∠DOC=2∠ACD=90°，
∴∠ACD=45°．
∴∠ACD+∠OCD=∠OCA=90°．
∵ point C is on circle o,
The straight line AC is the tangent line of circle o
(2) Method 1: ∵ od = OC = 2, ∵ doc = 90 °,
∴CD=2
2．
∵∠ACB=75°，∠ACD=45°，
∴∠BCD=30°，
When de ⊥ BC is used at point E, then ⊥ Dec = 90 °,
∴DE=DCsin30°=
2．
∵∠B=45°，
∴DB=2．
Method 2: connect Bo
∵∠ACB=75°，∠ACD=45°，
∴∠BCD=30°，∴∠BOD=60°
∵OD=OB=2
The △ BOD is an equilateral triangle
∴BD=OD=2．

In this paper, we prove that △ AC is a point in a graph

Proof: ab = AC,
∴∠ABC=∠ACB．
∵BD=CD．
∴∠1=∠2．
∴∠ABC-∠1=∠ACB-∠2．
That is ∠ abd = ∠ ACD

Circle O is the circumscribed circle of triangle ABC. The bisector ce of angle ACB intersects AB at point D, circle O at point E, and tangent EF of circle O intersects CB

Link be, CE bisection ∠ ACB ‰
∠ACE=∠BCE⇒AE=BE
∠ADE=∠BEC+∠EBA
∠EBA=∠ECA=∠ECB
∠EBF=∠BEC+∠ECB
So ∠ ade = ∠ EBF
EF is the tangent line {bef = ∠ EAD
⇒∆AED∽∆EFB
So AE · be = EF · ad AE = be
So AE2 = EF · ad

Let ABC be an extension of the circle a C = 3, and let C be the root of the circle

According to the chord tangent theorem:
∠BCD=∠A,
And ∠ d = ∠ D
So △ BCD ∽ CAD
So BC / AC = CD / ad = BD / CD
CD ^ 2 = BD * ad = BD * (AB + BD)
CD=2√3 AB=BC=4
So BD = 2
Therefore, according to BC / AC = BD / CD, AC = 4 √ 3

Circle O is the circumscribed circle of triangle ABC, ab = AC, passing through point a as PA, parallel to BC, intersecting extension line of Bo at point P. it is proved that AP is tangent of circle o

∵ circle O is the circumscribed circle of triangle ABC, ab = AC, passing through point a as PA, parallel to BC, intersection of extension line of Bo at point P  B = ∠ C
Ao bisection ∠ a ∠ Pao = ∠ CaO + ∠ PAC = 1 / 2 ∠ a + ∠ C = 90 ° AP is the tangent line of circle o

Circle O is the circumscribed circle of triangle ABC, and angle B = angle. CAD proves that ad is tangent of circle o The circle O is the circumscribed circle of the triangle ABCD, and the angle B = angle CAD. It is proved that ad is the tangent of circle o 2009-12-17 20:08 questioner: 410868313 | Views: 817 The circle O is the circumscribed circle of the triangle ABCD, and the angle B = angle CAD, Prove that ad is tangent of circle O The conditional angle B = angle CAD is changed to extend the intersection line ad of BC to D, and ad ^ 2 = DC * dB, other conditions remain unchanged, and whether AD is tangent of circle o

1) If you do the diameter ad of oand connect CE, then ∠ ace = 90 ° (the circumference angle of the diameter of the diameter), that is ∠ e ＋ ∠ EAC = 90 ° because ∠ e = ∠ B, (the circumference angle of the same arc) · ∠ B = CAD, so ∠ CAD + EAC = 90 °, that is, EA is perpendicular to ad, AE is the diameter of circle O, so ad is the tangent line of circle O. 2), make the diameter of O, AE is the tangent line of circular O. 2), make the diameter of O, connect EC, connect EC, in △ DAC and △ DAB ∠ D is common, AD / / dcbd = dcbd, AD / BD = DC / BD = DC / BD = DC / BD = DC / BD = DC / BD = DC / AD, So △ DAC ∽ DBA, ﹤ DAC = ∠ B. imitate 1) it can be proved that ad is tangent of circle o