Find a proof question: AB is the diameter of circle O, CD is a chord in the circle, AE ⊥ CD is in E, BF ⊥ CD is in F, and CE = DF

Find a proof question: AB is the diameter of circle O, CD is a chord in the circle, AE ⊥ CD is in E, BF ⊥ CD is in F, and CE = DF

yshyhua ,
It is proved that O is og ⊥ CD is g
∵AE⊥CD,BF⊥CD,OG⊥CD
∴AE‖OG‖BF
∵OA＝OB
∴EG/FG＝OA/OB＝1
∴EG＝FG
∵ og ⊥ CD, CD as string
∴CG＝DG
∵EG＝FG
∴CE＝DF

Given the diameter of ⊙ o AB = 20, chord CD intersecting AB at point G, Ag ⊥ BG, CD = 16, make AE ⊥ CD in E, BF ⊥ CD in F, then ae-bf=______ .

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As shown in the figure, the diameter of ⊙ o is ab = 2, am and BN are its two tangent lines, de cuts ⊙ o in E, intersects am in D, intersects BN in C. let ad = x, BC = y (1) Confirmation: am ‖ BN; (2) Find the relation between Y and X; (3) Find the area s of quadrilateral ABCD and prove that s ≥ 2

(1) It is proved that: AB is the diameter, am and BN are tangent lines,
∴AM⊥AB，BN⊥AB，
∴AM∥BN．
(2) If the crossing point D is DF ⊥ BC at F, then ab ∥ DF
From (1) am ∥ BN,  quadrilateral ABFD is a rectangle
∴DF=AB=2，BF=AD=x．
∵ De, Da, CE, CB are tangent lines,
According to the tangent length theorem, de = Da = x, CE = CB = y
In dfy + DC = 2, dfy = DC + C, dfy = dfy + C = dfy + C, dfy = dfy + C, dfy = dfy + C, dfy = dfy + C, dfy = dfy + C, dfy = dfy + C, dfy = dfy + C, dfy = dfy + C, dfy = DC,
∴（x+y）2=22+（y-x）2，
Simplify, y = 1
x（x＞0）．
(3) From (1) and (2), the area of quadrilateral s = 1
2AB（AD+BC）=1
2×2×（x+1
x），
That is, s = x + 1
x（x＞0）．
∵（x+1
x）-2=x-2+1
x=（
X-1
x） 2 ≥ 0, when and only when x=1, the equal sign holds
∴x+1
2, X ≥ 2

As shown in the figure, the diameter of circle O is ab = 12cm, am and BN are its two tangent lines, de tangent circle O to e, intersection am to D, intersection BN to C, let ad = x, BC = y When x is the value, the angle BCD is 60 degrees

 
It can be seen from the figure that when ∠ BCD = 60 °, the ∠ BCO = 30 °
According to the theorem of "the right angle side opposite to 30 ° is equal to half of the inclined side", the result is: CO = 2BO = 12cm
In addition, if de = ad, CE = BC, then CD = x + y
According to Pythagorean theorem, y = √ (12? 6? 2) = 6 √ 3,
OD = √（6² + X²）= √（36 + X²）
According to the theorem of "the right angle side opposite to 30 ° is equal to half of the slanted side", CD = 2 √ (36 + x 2) is obtained
Then x + 6 √ 3 = 2 √ (36 + x 2)
X = 2 √ 3
When x = 2 √ 3, ∠ BCD = 60 °

The diameter of circle O is ab = 12cm, am and BN are its two tangent lines, de tangent circle O to e, intersect am to D, intersect BN to C, let ad = x, BC = y, find the function of X and y The diameter of circle O is ab = 12cm, am and BN are its two tangent lines, de tangent circle O to e, intersect am to D, intersect BN to C, let ad = x, BC = y, find the functional relationship between X and y,

If AB is the diameter, am and BN are tangent lines, then am ⊥ AB; BN ⊥ ab
If DH is perpendicular to h, then DH = AB = 12; HC = BC-AD = y-x
If DC is tangent to circle O, then DC = de + DC = AD + BC = x + y
∵ DH ^ 2 + HC ^ 2 = DC ^ 2, i.e. 12 ^ 2 + (Y-X) ^ 2 = (x + y) ^ 2
∴xy=36

As shown in the figure, AB is the diameter of circle O, am and BN are its two tangent lines, de tangent circle O to point E, intersection am to point D, and intersection BN to point C,

The problem is: 1) proof: od is parallel to be 2) guess: what is the quantitative relationship between of and CD? And explain the reasons. (1) prove that: connect OE, ∵ am, de are tangent lines of ⊙ o,

As shown in the figure: ⊙ o diameter AB = 12, am and BN are its two tangent lines, de cut ⊙ o in E, intersect am in D, intersect BN in C, set ad = x, BC = y, find the functional relationship between Y and X, and draw its general picture

DF ⊥ CB is made through D, and CB is crossed at point F,
∵ DA and DC are tangent lines of circle o,
∴DA=DE，
CB and CE are tangent lines of circle o,
∴CB=CE，
And ∠ DAB = ∠ ABF = ∠ BFD = 90 °,
The quadrilateral ABFD is rectangular,
∴DA=FB，DF=AB，
In a right triangle CDF,
∵AD=x，BC=y，AB=12，
∴CD=CE+ED=DA+CB=x+y，DF=AB=12，CF=CB-FB=y-x，
According to Pythagorean theorem, CD2 = df2 + CF2,
That is (x + y) 2 = 122 + (Y-X) 2,
The result is: xy = 36, that is, y = 36
x（x＞0）；
Draw the function image in the plane rectangular coordinate system, as shown in the figure

In the circle O, the chord AB intersects OC and OD respectively at point M. if n ∠ AMC = ∠ Bnd, it is proved that am = BN

Join OA, ob, triangle oan and triangle OBM congruent, an = BM (common part Mn), and subtract Mn on both sides to obtain am = BN

As shown in the figure, AB is the diameter of circle O, m and N are the midpoint of AO and Bo, cm ⊥ AO and DN ⊥ ob respectively. It is proved that AC = BD =Why is another question always submitted automatically

∵OC=OD= r/2,OM=ON
∴RT△OCM≌RT△ODN （HL）
∴CM=DN
∵AM=BN,∠CMA=∠DNB=90°
∴△AMC≌△BND
∴AC=BD

As shown in the figure, the straight line AB and ⊙ o with radius of 2 are tangent to point C, D is the point on ⊙ o, and ∠ EDC = 30 °, chord EF ∥ AB, then the length of EF is () A. 2 B. 2 Three C. Three D. 2 Two

OE and OC are connected, and the intersection point of OC and EF is m
∵∠EDC=30°，
∴∠COE=60°．
∵ AB and ⊙ o are tangent,
∴OC⊥AB，
And ∵ EF ∫ ab,
⊥ EF, that is, △ EOM is a right triangle
In RT △ EOM, EM = sin60 °× OE=
Three
2×2=
3，
∵EF=2EM，
∴EF=2
3．
Therefore, B