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As shown in the figure, in the rectangle ABCD, e is the midpoint of BC, and de ⊥ AC, then CD: ad=______ .
As shown in the figure: ∵ ad ∥ BC, e is the midpoint of BC,
∴△ECO∽△DAO，
∵AD=BC，EC=1
2BC
∴EC
AD＝ CO
AO=1
2；
∵ ADC = 90 ° AC ⊥ ed, ∵ CAD is the common angle of △ ADC and △ AOD,
∴△ADC∽△AOD，
Similarly, △ ADC ∽ doc can be obtained,
﹤ ADC ∽ AOD ∽ doc, namely ad
DC＝ AO
OD＝ DO
OC，
∵ it has been proved that CO: Ao = 1:2,
∴OD=
AO•OC=
2, namely CD: ad=
2：2．
So the answer is:
2：2
In circle O, chord AB = AC, angle BAC = 60 degrees, D is any point on arc BC, if ad = 2, find the area of quadrilateral ABCD
Let d be on the extension line of Ao, so ad bisects the angle BAC,
So the angle bad is 30 degrees
Because ad = 2
So BD = 1
AB = radical 3
So the s triangle abd = root 3 / 2
Similarly, s triangle ADC = root 3 / 2
So the s-quadrilateral ABCD = radical 3
Given that AB is the diameter of a circle, ab = 10, chord ad = 6, and point C is a moving point on BD arc, the maximum value of ABCD area of quadrilateral is calculated?
The area of the triangle abd is fixed, and the key is the triangle BDC, where BD is a fixed length. When the distance between C and BD is the largest, the area will be the largest. Obviously, when CO is perpendicular to BD, it is satisfied. (o is the center of the circle)
If the circle O is circumscribed to the square ABCD with side length 2, P is a point on the arc ad, and AP = 1, then (PA + PC) △ Pb= I still look tired. Can you give me more details
AP = 1, AC = 2 √ 2, PC = √ 7
Let Pb and ad intersect with M. △ APM ∽ MBP.PM /DM =AM /BM =1/（2√2）
PM / am = DM / BM = 1 / (2 √ 2) = Pd / AB, PD = √ 2 / 2, △ PBD
PB=√30/2,(PA+PC)÷PB=（1+√7）/√30/2
As shown in the figure, it is known that square ABCD is the inscribed square of circle O, and E, F, G and H are the midpoint of arc AD, arc DC, arc CB and arc BA respectively. Verification: eight sided AHBG Cfde is a regular octagon Tonight,
The square ABCD is the inscribed square of circle o
So AB = BC = CD = da
E. F, G and H are the midpoint of arc ad, arc DC, arc CB and arc Ba respectively
So,
AE=DE=DF=CF=CG=BG=DH=AH
Therefore, the octagon ahbgcfde is a regular octagon
As shown in the figure, AB is the diameter of ⊙ o, the degree ratio of quadrilateral ABCD connected to ⊙ o, arc BC, arc CD, arc ad is 3:2:4, Mn is the tangent of ⊙ o, C is the tangent point, then the degree of ⊙ BCM is______ Degree
When OC is connected, the degree ratio of arc BC, arc CD and arc ad is 3:2:4; if BC = 3x, then CD = 2x, ad = 4x, ∵ BC + CD + ad = 180 °, that is, 3x + 2x + 4x = 180 °, x = 20 °, 3x = 60 °, i.e., ∵ BOC = 60 °, ∵ ob = OC, ∵ OBC = ∵ OCB = 12 (180 ° - ∠ BOC) = 12 (
As shown in the figure, take the vertex a of the parallelogram ABCD as the center of the circle, and ab as the radius to make the circle, make ad, BC at e, F, and extend the intersection of BA ⊙ a to g GE＝ EF．
Proof: connect AF,
∵AB=AF，
∴∠ABF=∠AFB．
∵ quadrilateral ABCD is a parallelogram,
∴AD∥BC．
∴∠DAF=∠AFB，∠GAE=∠ABF．
∴∠GAE=∠EAF．
Qi
GE＝
EF．
1. As shown in the figure, in the isosceles trapezoid ABCD, ad ∥ BC, ab = DC = 5, ad = 2, BC = 8, the vertex e of ∥ men moves on BC. One edge always passes through point a, and the other side intersects with point F with CD, so that ∠ men = ∠ B, connect AE. If △ AEF is an isosceles triangle (AF = EF), find the length of be 2. Given am ‖ BN, ∠ a = ∠ B = 90 ° AB = 4, point D is a moving point on the ray am (not coincident with a), point E is a moving point on line AB (not coincident with a and b), connecting De, passing through point E as the vertical line of De, intersection line BN at point C, let AE = x, BC = y, if ad = de = AB, is C △ BCE a fixed value?
(1) ∵ AB = DC = 5, ∵ B = C (1 point) and AEC = ∠ B + ∠ BAE = ∠ AEF + ∠ FEC ? AEF = ∠ B, ∵ BAE = ∵ FEC (1 point) ∵ Abe ∵ ECF (1 point) ∵ AB be = EC FC, i.e. 5 x = 8-x 5-y (1 point) ? y = 1 5 (x2-8x + 25) (0 ≤ x ≤ 8) (2 points) (2) pass through a and
As shown in the figure, in the isosceles trapezoid ABCD, ad ∥ BC, ab = CD, diagonal AC ⊥ BD, ad = 4cm, BC = 10cm, calculate the area of trapezoid ABCD
D is used as de ∥ AC, the extension line of BC is at e, and over D is DF ⊥ BC at F
∵AD∥CB，DE∥AC，
The quadrilateral Adec is a parallelogram,
∴DE=AC，AD=CE=4
∵ in the isosceles trapezoid ABCD, ab = CD,
∴DE=AC=BD，
∵AC⊥BD，CE∥AD，
∴DE⊥BD，
The △ BDE is an isosceles right triangle,
And ∵ ad = 4, BC = 10,
∴DF=1
2BE=1
2（AD+BC）=1
2（4+10）=7cm，
The area of trapezoid is: 1
2（4+10）×7=49．
So the answer is: 49
Given that the vertices of the isosceles trapezoid ABCD are all on ⊙ o, ab ∥ CD, arc AB + arc CD = arc AD + arc BC, if AB = 4, CD = 6, calculate the area of the isosceles trapezoid ABCD
If the center of the circle is O and the radius is r, then
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