As shown in the figure, △ ABC, e is the heart of △ ABC, the bisector of ∠ A and the circumscribed circle of △ ABC intersect at point D. It is proved that de = dB

As shown in the figure, △ ABC, e is the heart of △ ABC, the bisector of ∠ A and the circumscribed circle of △ ABC intersect at point D. It is proved that de = dB

Connect be,
∵ e is the heart,
ν AE, be are the angular bisectors of ∠ BAC, ∠ ABC,
∴∠BED=∠BAE+∠EBA，∠EBA=∠EBC，∠BAE=∠EAC，
∴∠BED=∠EBC+∠EAC，∠EBD=∠EBC+∠CBD，
A kind of
CD=
CD，
∴∠EAC=∠CBD，
∴∠EBD=∠BED，
∴DE=BD．

As shown in the figure, the triangle ABC is inscribed in the circle O, AE is the diameter of circle O, ad is perpendicular to point D, AE is the diameter of circle O, verification: ab × AC = ad × AE

prove:
Link be
∵ AE is the diameter
∴∠ABE=90°
∵AD⊥BC
∴∠ADC=90°
∴∠ABE=∠ADC
∵∠E=∠C
∴△ABE∽△ADC
∴AB/AD=AE/AC
∴AB×AC=AD×AE

It is known that the extension lines of chords AB and CD of circle O intersect at point P. Po is the bisector of angle APC, and points m and N are the midpoint of arc AB and arc CD respectively. It is proved that Mn is perpendicular to Po

Connect OM and on to AB, CD to e and f respectively
M. N is the midpoint of arc AB and arc CD respectively, and OM and on are the vertical bisectors of AB and CD respectively
That is, OE ⊥ AB, of ⊥ CD
∵ Po is the bisector of ∵ APC
∠POE=∠POF
In △ omn, OM = on, that is, △ omn is an isosceles triangle and ∠ Poe = ∠ POF
OP is the bisector of mon
Then OP is △ omn is the height on the bottom edge of isosceles triangle Mn
That is op ⊥ Mn

The extension line of diameter ab of circle O intersects with the extension line of chord CD at point P, e is a point on the circle, arc AE = arc AC, de intersects AB at point F. it is proved that PF * Po = PA * Pb Ten minutes, quick

Proof: connect AE, AC,
Because the angle PDF is an outer corner of the circle inscribed quadrilateral AEDC,
So angle PDF = angle EAC,
Because arc AE = arc AC, AB is the diameter of circle o,
So arc EB = arc CB, that is arc EBC = 2 arc CB,
So the circumference angle EAC = the center angle cob,
So angle PDF = angle cob,
Because angle P = angle P,
So triangle PDF is similar to triangle POC,
So Pd / PO = PF / PC,
So pf * Po = PC * PD
The secant theorem shows that PC * PD = PA * Pb,
So pf * Po = PA * Pb

Known AB、 CD is two arcs of the same circle, and AB=2 CD, then the relationship between chord AB and 2CD is () A. AB=2CD B. AB＜2CD C. AB＞2CD D. Not sure

As shown in the figure, if the Arc de = arc CD is intercepted on the circle, then arc AB = arc CE
∴AB=CE
∵CD+DE=2CD＞CE=AB
∴AB＜2CD．
Therefore, B

Known: as shown in the figure, in ⊙ o, the chord AB = CD It is proved that: (1) arc AC = arc BD; （2）∠AOC=∠BOD．

It is proved that: (1) in ⊙ o, the chord AB = CD,
/ / arc AB = arc CD,
∵ arc BC = arc CB,
ν arc AC = arc BD;
(2) ∵ arc AC = arc BD,
∴∠AOC=∠BOD．

As shown in the figure, in circle O, arc AC = arc BD, then explore the quantitative relationship between chord AB and CD, angle AOC and angle BOD

∵ arc AC = arc BD,
﹤ AOC = ∠ BOD (in the same circle, the center angle of the circle opposite to the equal arc is equal)
String ab ‖ CD or AB = CD

In circle O, if arc AB = arc 2CD, then the relationship between chord AB and CD is

CD As shown in the figure, arc CD = arc BD, so arc CD = 1 / 2 arc ab
Because CD + BD > AB in △ ADB, AB < 2CD

In circle O, the string AB is parallel to the string CD. How to prove that arc AC = arc BD

It is proved that if we connect ad angle ADC = angle bad (AB parallel CD, equal staggered angle), then arc AC = arc BD (in the same circle, the circumference angle is equal, the opposite arc is also equal)

AB is the diameter of circle O, chord CD intersects AB at point m, and OM = cm. Try to determine the quantitative relationship between arc BD and arc AC, and explain the reasons

Drawing, connecting OC, extending OC intersection circle e, connecting AE EB AC CB, angle DCE equals angle EOB equals 2ecb, angle EAB equals ABC equals ECB equals 2 / 3dcb, so arc AC 2 / 3 arc dB