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As shown in the figure, AB is the diameter of circle O, CB is the chord, OD ⊥ CB is in E, cross arc CB is in D, and ac. (1) please write down two different types of As shown in the figure, AB is the diameter of circle O, CB is the chord, OD ⊥ CB is in E, cross arc CB is in D, and AC is connected (1) Please write two different types of correct conclusions; (2) If CB = 8, ed = 2, find the radius of circle o
1. Conclusion: 1) AC
As shown in the figure, O is the center of the circle, the radius OC is the vertical chord AB, the perpendicular foot is the point D, OC = 5, ab = 8, find the length of OD
Where is the picture?
Connect OA and set it to X
OD=X-CD
In △ oad, according to Pythagorean theorem, x = 5 od = 3 of X is obtained
It is known that AB is the chord of ⊙ o, and the straight line of radius od is perpendicular to C, if AB = 2 3 cm, OC = 1 cm, find the length of CD
As shown in the figure, connect OA
∵ AB is the chord of ⊙ o, the straight line of radius od is perpendicular to C, ab = 2
3 cm,
∴AC=1
2AB=
3 cm
And ∵ OC = 1 cm,
In the right angle △ AOC, the result of Pythagorean theorem is: OA=
Ac2 + oc2 = 2cm,
ν CD = oa-oc = 1cm
In 0 o, if the chord AB = 8, the radius OC ⊥ AB is in D, and OD = 2CD, find the length of OC
It is obvious to use Pythagorean theorem to solve it
Connect OA
If CD = x, then od = 2x, OC = OA = 3x
In RT △ ADO, ad = 1 / 2Ab = 4. According to Pythagorean theorem (2x) ^ 2 + 4 ^ 2 = (3x) ^ 2
The solution is x = 4 (Radix 5) / 5
So OC = 3x = 12 (Radix 5) / 5
If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen
O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C
AB is the diameter of circle O, BC is the chord of circle O, OD is perpendicular to CB at point C, intersect arc BC at point D, connect CD, find out a relationship between angle CDB and angle ABC and prove it
A: the relationship between angle CDB and angle ABC is: ∠ CDB = ∠ ABC + 90,
Because ∠ ABC = ∠ ABC (equal to the circumference angle of the arc)
∠ ADB = 90 degrees
So: ∠ CDB = ∠ ABC + 90,
As shown in the figure, ad is the middle line of △ ABC, be intersects AC to e, and ad to F, and AE = EF. Verification: AC = BF
It is proved that ∵ ad is the midline of ᙽ ABC,
∴BD=CD．
Method 1: extend ad to point m, make MD = FD, connect MC,
In △ BDF and △ CDM,
BD＝CD
∠BDF＝∠CDM
DF＝DM
∴△BDF≌△CDM（SAS）．
∴MC=BF，∠M=∠BFM．
∵EA=EF，
∴∠EAF=∠EFA，
∵∠AFE=∠BFM，
∴∠M=∠MAC，
∴AC=MC，
∴BF=AC；
Method 2: extend ad to point m, make DM = ad, connect BM,
In △ ADC and △ MDB,
BD＝CD
∠BDM＝∠CDA
DM＝DA ，
∴△ADC≌△MDB（SAS），
∴∠M=∠MAC，BM=AC，
∵EA=EF，
Ψ cam = ∠ AFE, and ∠ AFE = ∠ BFM,
∴∠M=∠BFM，
∴BM=BF，
∴BF=AC．
It is known that: as shown in the figure, D is a point on the BC side of △ ABC, e is a point on ad, EB = EC, ∠ Abe = ∠ ace, and the verification is: ∠ BAE = ∠ CAE
Proof: EB = EC,
∴∠EBD=∠ECD，
And ∵ Abe = ∠ ace,
∴∠ABC=∠ACB，
∴AB=AC，
In △ Abe and △ ace
AB＝AC
EB＝EC
AE＝AE
∴△ABE≌△ACE，
∴∠BAE=∠CAE．
It is known that: as shown in the figure, D is a point on the BC side of △ ABC, e is a point on ad, EB = EC, ∠ Abe = ∠ ace, and the verification is: ∠ BAE = ∠ CAE
Proof: EB = EC,
∴∠EBD=∠ECD，
And ∵ Abe = ∠ ace,
∴∠ABC=∠ACB，
∴AB=AC，
In △ Abe and △ ace
AB＝AC
EB＝EC
AE＝AE
∴△ABE≌△ACE，
∴∠BAE=∠CAE．
It is known that: as shown in the figure, D is a point on the BC side of △ ABC, e is a point on ad, EB = EC, ∠ Abe = ∠ ace, and the verification is: ∠ BAE = ∠ CAE
Proof: EB = EC,
∴∠EBD=∠ECD，
And ∵ Abe = ∠ ace,
∴∠ABC=∠ACB，
∴AB=AC，
In △ Abe and △ ace
AB＝AC
EB＝EC
AE＝AE
∴△ABE≌△ACE，
∴∠BAE=∠CAE．
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