As shown in the figure, in the circle O, the chord AB intersects OC, OD at m, n respectively. If AMC = Bnd, it is proved that am = BN

It is proved that O is OE ⊥ AB, and the vertical foot is e,
Because ∠ AMC = ∠ Bnd
∠AMC=∠OMN
∠BND=∠ONM
So ∠ omn = ∠ Onm
So om = on
So me = ne, (three wires in one)
Because OE ⊥ ab
So AE = be (vertical diameter theorem)
So ae-me = be-ne
BN am = BN am

It is known that, as shown in the figure, AB is the radius of circle O, chord CD ∥ AB, straight lines cm and DN cut semicircles at points c and D, and intersect with lines AB at points m and N respectively (1) Results: Mo = No; (2) Let the angle M = 30 ° and prove that Mn = 4CD

(1) In this paper, we make a study of the linear cm and DN, and cut the semicircle to the point C, D  OCM = ∵ OCM = ∵ ODN = 90 ° and ∵ OC = OD,  OCM 8780\8780\\8780\\87808780878087808780\878787878780878787878787878787878787878787 \9 com = arc AC = arc AC = arc degrees

In the circle O, the chord AB is perpendicular to the chord CD in M. the radius of circle O is known to be 5, BM = 6, am = 2. Find the size of dm-cm

For OE, if AB is perpendicular to e, be = BA / 2 = (6 + 2) / 2 = 4; if ob is connected, OE = √ (OB ^ 2-BE ^ 2) = √ (25-16) = 3. If CD is perpendicular to F, then CF = CD / 2. ? OEM = ∠ EMF = ∠ MFO = 90 °. Quadrilateral OEMF is rectangular, FM = OE = 3. So dm-cm = FM + df-cm = FM + cf-cm = FM + FM = 6

As shown in the figure, AB is the diameter of ⊙ o, chord CD ⊥ AB is at point m, am = 2, BM = 8, find the length of CD

Connect OC,
⊥ is the diameter of ab ⊙ M,
∴CD=2CM，
∵AM=2，BM=8，
∴AB=10，AC=AO=5，OM=AO-AM=3，
In RT △ CMO, CM=
CO2−OM2=4，
∴CD=8．

Known: as shown in the figure, the diameter PQ of ⊙ o intersects chord AB and CD respectively at points m, N, am = BM, ab ∥ CD Verification: DN = CN

It is proved that ∵ PQ is the diameter, am = BM,
ν PQ ⊥ AB in M
And ∵ ab ∥ CD,
⊥ CD in n
∴DN=CN．

As shown in the figure, AB is the chord of circle O, CD tangent circle O to point m, and CD ∥ AB, prove that am = BM

Because CD is tangent to point m, CD ⊥ OM, because CD ∥ AB, so ab ⊥ OM, then ⊥ MNA and triangle MNB are congruent, so am = BM

As shown in the figure, M is a point in circle O. use ruler to make a chord so that ab passes through point m and am = BM

The main process is divided into two steps: (1) determine the center of the circle: take three points n, P, Q randomly on the circumference to make Mn, MP vertical bisector. The specific operation is as follows: take n, P as the center of the circle, the length greater than NP / 2 is the radius, the two arcs intersect at two points, making a straight line through these two points is NP vertical bisector, taking n, Q as the center of the circle, and the length greater than NQ / 2 as the radius

It is known that in △ ABC, ab = AC, O is a point in △ ABC, and the extension line of OB = OC, Ao intersects BC at D.

prove:
∵AB=AC,OB=OC,AO=AO
∴△AOB≌△AOC
∴∠BAO=∠CAO
∵AB=AC
⊥ BC, BD = DC (isosceles triangle with three lines in one)

As shown in the figure, △ ABC, ab = AC, O is a point in △ ABC, and ∠ OBC = ∠ OCB

It is proved that: ∵ AB = AC,  ABC = ∠ ACB (equilateral and equal angle),
∵∠OBC=∠OCB，
Ψ ABO = ∠ ACO, OB = OC (equiangular to equilateral),
∴△AOB≌△AOC（SAS），
∴∠OAB=∠OAC，
And ∵ AB = AC,
⊥ BC

It is known that in △ ABC, ab = AC, O is a point in △ ABC, and the extension line of OB = OC, Ao intersects BC at D. ; verification: ad ⊥ BC, BD = CD

prove:
∵AB=AC,OB=OC,AO=AO
∴△AOB≌△AOC
∴∠BAO=∠CAO
∵AB=AC
⊥ BC, BD = DC (isosceles triangle with three lines in one)

As shown in the figure, in the circle O, the chord AB intersects OC, OD at m, n respectively. If AMC = Bnd, it is proved that am = BN