![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Given that ∠ AOB = 90 degrees, D, C divide AB into three equal parts, chord AB intersects with radius OC, OD at points E and F, and proves that AE = DC = BF
C divide AB into three equal parts
Then ∠ AOC = ∠ cod = ∠ DOB, AC = DC = BD
The circular angle of chord BC is ∠ BAC, and its center angle is ∠ BOC
Then ∠ BAC = ∠ BOC / 2 = (∠ cod + ∠ BOD) / 2 = ∠ cod = ∠ AOC
Because the radius OA = OC
There is ∠ OAC = ∠ OAB + ∠ BAC
∠AEC=∠OAB+∠AOC
So ∠ OAC = ∠ AEC
Because the radius OA = OC
So ∠ OAC = ∠ OCA
That is, ∠ OCA = ∠ AEC
So AE = AC
Similarly, BF=DB
So AE = DC = BF
As shown in the figure, AC and BD intersect at point O, and ab ‖ DC, OA = ob Confirmation: OC = OD
Proof: Ao = Bo,
∴∠A=∠B，
∵DC∥AB，
∴∠D=∠B，∠C=∠A，
∴∠C=∠D，
∴CO=DO．
3. As shown in the figure, OC and OA on both sides of the rectangular aocb are located on the x-axis and y-axis respectively. The coordinates of point B are B (- 20 / 3,5), and D is a point on AB side. Fold the triangle ADO along the straight line od so that point a falls on point E on the diagonal ob. if point E is on the image of an inverse scale function, then what is the analytic formula of the function? If the inverse proportional function and ab intersect at M and BC intersect n, find s △ mon
According to the meaning of the question, | Ao | = 5, | ab | = 20 / 3, therefore, | Bo | = 25 / 3 according to Pythagorean theorem, and | ad | = 5 / 2 are obtained according to the angle bisector theorem
As shown in the figure, OC and OA on both sides of the rectangular aocb are located on the x-axis and y-axis respectively, and the coordinates of point B are b-20 3,5), D is a point on the edge of AB, fold △ ADO along the straight line OD, so that point a just falls at point E on the diagonal ob. If point E is on the image of an inverse scale function, then the analytic formula of the function is () A. y=12 X B. y=6 X C. y=-6 X D. y=-12 X
According to the Pythagorean theorem, OB = 52 + (203) 2 = 253, ∵ △ OEF ∽ OBC, ∵ EFBC = oeob, that is, EF5 = 5253. The solution is: EF = 3, and ∵ the coordinates of point a
As shown in the figure, the length OA of the rectangular oabc is 3, and the width OC is 1. Fold △ AOC along AC to obtain △ APC. (1) fill in the blank: ∠ PCB=____ The coordinates of point P are
(1) Fill in the blank: ∠ PCB=__ 30__ The coordinates of point P are (root of 2 / 3,3 / 2)
As shown in the figure, in the plane coordinate system, points a and B are on the x-axis and y-axis respectively, OA: OB = 1:2, C is the midpoint of the line segment, and OC = 3 times the root sign 5, As shown in the figure, in the plane coordinate system, points a and B are on the x-axis and y-axis respectively, OA: OB = 1:2, C is the midpoint of the line segment, and OC = 3 times the root sign 5, point D is on the line OC, and the abscissa of point D is 2 (1) Find the length of OA and ob 2. Find the intersection of line AB and Y axis and point E, and find the area of quadrilateral cdeb 3. If the line ad intersects the Y-axis and the point E, find the quadrilateral cdeb 4. P is the point on the straight line ad. is there a point Q in the plane so that Q, a, P, O are the vertices of the quadrilateral is a diamond?
Let | OA | = m, | ob | = n
According to the title: m ^ 2 + n ^ 2 = 45
n=2m
The solution is m = 3, n = 6
∴ |OA|=3 |OB|=6
Let AB pass through the first, second and fourth quadrants
Then point a coordinates (3,0) B (0,6)
Then the coordinates of the midpoint C (3 / 2,3)
Equation y = 2x for line OC
When x = 2, y = 4
The coordinates of point D are (2,4)
The slope of straight line ad is k = (4-0) / (2-3) = - 4
Equation y-0 = - 4 (x-3) of straight line ad
y=-4x+12
The coordinates of the point e of the line on the y-axis are (0,12)
S quadrilateral cdeb = s △ ode-s △ OBC
=(1/2)×|OE|×2-(1/2)×(3/2)×|OB|
The abscissa of the abscissa of 2 is the abscissa of 2
=(1/2)×(12)×2-(1/2)×6×3/2
=12-4.5
=7.5
3. A: there is
Because the diagonals of the doves are perpendicular to each other,
The point P should be on the vertical bisector of OA
The abscissa of P is 3 / 2
Put the abscissa into the equation of AD
y=-4x+12=-4×3/2+12=6
The coordinate of P is (3 / 2,6), while that of point q is (3 / 2, - 6)
It can be calculated that when the line AB passes through one, three and four quadrants, the situation is exactly the same
The coordinates of point P at this time are the coordinates of point Q now
The problem has been solved. Is it complete? Can you give a score?
As shown in the figure, in the plane rectangular coordinate system xoy, the two sides of the rectangular oabc are respectively on the x-axis and y-axis, OA = 10cm, OC = 6cm As shown in the figure, in the plane rectangular coordinate system xoy, the two sides of the rectangular oabc are respectively on the x-axis and y-axis, OA = 10cm, OC = 6cm. The existing two moving points P and Q start from O and a at the same time Question (2) "set the velocity of point Q as..." A detailed explanation of?
Problem: as shown in the figure, in the plane rectangular coordinate system, the two sides of the rectangular oabc are on the x-axis and y-axis respectively, OA = 10cm, OC = 6cm. P is the moving point on the line OA. Starting from point O, moving uniformly along the direction of OA at the speed of 1cm / s, the point q is on the line ab. it is known that the distance between a and Q is a times of the distance between O and P
In the plane rectangular coordinate system, the two sides of the rectangular oabc are on the x-axis and y-axis respectively, OA = 8 times the root sign 2cm, OC = 8cm. The existing two moving points P and Q are the same from O and C respectively
It is suggested that we should check whether the question is complete or not, otherwise it is impossible to get the answer
In the plane rectangular coordinate system, the two sides of the rectangular oabc are on the x-axis and y-axis respectively, OA = 8, root sign 2cm, OC = 8. There are two moving points PQ, starting from O and C, P moves along OA direction at the speed of root sign 2cm per second, Q moves 1cm per second on CO, and the time is T seconds, 1. The area of triangle OPQ is expressed by the formula containing t 2. Verification: the area of quadrilateral opbq is a fixed value, and the fixed value is obtained Complementary graphs with graphs
One
OC= 8-t
OQ= 2t
The area of triangle OPQ = 1 / 2 * OQ * OP = 1 / 2 * (8-radical 2 * t) * 2T = (8-radical 2 * t) t
Two
The area of quadrilateral opbq = the area of triangle OBQ + the area of triangle OBP
= 1/2*OQ*BC+1/2*OP*AB
=8 radical 2 * 1 / 2 * (8-radical 2 * t) + 2T * 8 * 1 / 2
=32 root number 2 + (8-4 root number 2) t
=32 root sign 2
As shown in the figure, in the plane rectangular coordinate system, the two sides of the rectangular oabc are on the x-axis and y-axis respectively, with OA = 10 cm and OC = 6 cm. P is the moving point on the line OA, Starting from point O, move uniformly along OA direction at the speed of 1cm / s. point q is on the line ab. it is known that the distance between points a and Q is a times of that between O.P. if (a, t) is used to represent the passing time t (s), two triangles in triangle OCP. Triangle PAQ and triangle CBQ are congruent. Please write all possible situations of (a, t)
Using t, a and known data to express the length of the following line segment is as follows: OC = AB = 6, OA = BC = 10, Op = t, AP = 10-T, AQ = at, BQ = 6-at, △ OCP ≌ △ Apq: OP = AQ, OC = AP. Substituting the above length, we can get (a, t) = (1,4) △ OCP ≌ △ AQP, Op = AP, OC = AQ, then (a, t) = (1.2,5)
RELATED INFORMATIONS
- 1. As shown in the figure, in the plane rectangular coordinate system, the two sides of the rectangular oabc are respectively on the x-axis and y-axis, OA = 10 cm, OC = 6 cm, and the existing two moving points P and Q are from the As shown in the figure, in the plane rectangular coordinate system, the two sides of the rectangular oabc are respectively on the x-axis and y-axis, OA = 10 cm, OC = 6 cm. The existing two moving points P and Q start from D and a at the same time, point P moves uniformly along the direction of OA on line OA, and point Q moves uniformly along AB direction on line ab. the velocity of point P is known to be 1cm / s (1) Suppose that the velocity of point q is cm / s and the movement time is T seconds, ① When the area of △ CPQ is the smallest, find the coordinates of point Q; ② when △ cop and △ PAQ are similar, find the coordinates of point Q (2) Suppose the velocity of point q is a cm / s, ask if there is a value of a, so that △ OCP is similar to △ PAQ and △ CBQ? If so, ask for the value of a and write out the coordinates of point Q at this time; if not, please explain the reason
- 2. In circle O, the chord Mn / / EF, P is a point on circle O. the radius of circle O is known to be 10cm, Mn = 12cm, EP = 16cm. Find the distance between chord Mn and ef
- 3. As shown in the figure, in ▱ ABCD, AE bisects ∠ bad to intersect DC at point E, ad = 5cm, ab = 8cm, and calculate the length of EC
- 4. If the lengths of two parallel chords are 6cm and 8cm respectively in a circle with a radius of 5cm, the distance between the two chords is______ .
- 5. It is known that, as shown in the figure, the diameter ab of the circle O intersects with the chord CD at the point E, AE = 1, be = 5, ∠ AEC = 30 ° and find the length of CD I can't draw a picture
- 6. 0
- 7. If AB is 10, Pb is 4 and OP is 5, then find the radius of the circle
- 8. It is known that in the triangle ABC of RT, the angle ACB = 90 degrees, D and E are the midpoint of AB and BC respectively, and the point F is on the extension line of AC, and CF = De
- 9. 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
- 10. In ⊙ o, the chord ab ⊥ CD is in M. the radius of ⊙ o is 5, BM = 6, am = 2. Find the size of dm-cm
- 11. As shown in the figure, OA ⊥ OC, ob ⊥ OD, ∠ BOC = 30 ° and find AOB, COD, AOC
- 12. In the circle O, AB is the diameter, the chord CD is perpendicular to P, AC = DC = 2 times the root sign 3, and find the length of Op
- 13. As shown in the figure, ad is the diameter of ⊙ o, AC is the chord, ∠ CAD = 30 °, ob ⊥ ad is in O, AC is in B, ab = 5, and the length of BC is obtained
- 14. Circle O1 and circle O2 intersect at two points a and B, P is the point on circle O1, connecting PA and Pb, intersecting circle O2 with C and D. It is proved that po1 is perpendicular to CD
- 15. As shown in the figure, point h is the vertical center of △ ABC, the circumscribed circle ⊙ O2 of ⊙ O1 and ⊙ BCH with ab as diameter intersect at point D, and extend ad to intersect ch at point P, Verification: point P is the midpoint of Ch
- 16. As shown in the figure, circle I is the inscribed circle of the triangle ABC, which is tangent to points D, e and F with AB, BC and Ca respectively, and the angle def = 50 degrees. Find the angle A
- 17. It is known that the radii of circle O1 and circle O2 are 3cm and 5cm respectively, and if they are inscribed, the position relationship between the center of circle and O1O2? What is the positional relation? Which is the extrinsic intersection and tangent? It is known that the radius of circle O1 and circle O2 are 3cm and 5cm respectively, and if they are inscribed, the center distance of circle is 5cm!
- 18. Given that circle O1 and circle O2 are inscribed at point a, the chord ab of circle O1 intersects with circle O2 at point C, the ratio of radius of circle O1 to circle O2 is 2:3, ab = 12 emergency
- 19. The third degree of the sixth root is not equal to 5:10 of the third part of the root sign
- 20. In a circle, if the chord AB = 1 / 3 of the root of 2 times and the chord center distance is 1, what is the radius of the circle? A, 2 B, 13 C, 11, D, 3 What to choose?