As shown in the figure, place a rectangle in the plane rectangular coordinate system, OA = 2, OC = 3, e is the midpoint of AB, and the image of inverse scale function passes through point E and connects with BC Intersection at point F (1) Finding the analytic formula of straight line OB and the analytic formula of inverse scale function (2) Connect of and OE, point P is on the straight line ob, and s triangle AOP = 2S quadrilateral oebf, calculate the coordinates of point P

We don't know how to place the rectangular oabc. Take a on the positive half axis of X axis and C on the positive half axis of Y axis as an example, the solution is. 1, from the topic a (2,0), B (2,3), C (0,3), O (0,0), E (2,3 / 2). Because ob passes ob, let y = KX, so k = 3 / 2, y = 3 / 2x. Because the inverse proportion function y = m / X passes e, so m = 3, y = 3 /

It is known that in the plane rectangular coordinate system as shown in the figure, the straight line y = 2 / 3 / - 2 / 3 intersects the edges OC and BC of rectangle ABCD at e and f respectively. It is known that Ao = 3. OC = 4 to find △ CEF

Let four vertices of a convex quadrilateral ABCD satisfy the following conditions: the sum of the distances from each point to the other three points must be equal. What kind of quadrilateral is it? Proof

From the perspective of point a, s = AB + AD + AC ① From the perspective of point B, s = Ba + BD + BC ② From the perspective of point C, s = Ca + CB + CD ③ From the perspective of point D, s = Da + DB + DC ④ AD + DC = AB + BC (1) From ① and ③, Ba + ad = CB + CD (2) From (1) + (2), 2ad = 2

Find a point O in the quadrilateral ABCD to minimize the sum of the distances to the four vertices of the quadrilateral, and please give your reasons

Intersection o of diagonal lines of four sides

Let the area of the square be 1, e and f be the midpoint of AB and AD & nbsp; respectively, and GC = 13fc, then the area of the shadow is______ ．

If the parallel line of AD and BC crosses AB to P and ad to Q, the shadow area is 12 × be × GP. Because be = 12ab, FG: GC = 2:1, QD = 13fd = 16ad, GP = AQ = 1-16 = 56ad, the shadow area is 12 × 12ab × 56ad = 524ab × ad = 524. A: the shadow area is 524

ABCD is a square with side length 4, E.F is the midpoint of AB and ad respectively, GC is perpendicular to the plane ABCD, and GC = 2, connect Ge and GF, and calculate the distance from point B to plane EFG

AC and BD intersect at O, EF and AC intersect at P, AC = 4 root sign 2; CP = 3 root sign 2 EF | BD, so BD | plane EFG. Therefore, the distance from point B and point O to plane EFG is equal in triangle ACG. If perpendicular OQ is made from point o to Ag, OQ = O, the distance from point to plane EFG is easy to get in triangle ACG and OPQ: OQ = (2 root sign 11) / 11, so the distance from point B to plane EFG = (2 root sign 11) / 11

Point E is the midpoint of CD side of parallelogram ABCD, connecting be and extending the extension line of intersection ad at point F. 1) e is the midpoint of BF
2) Angle f = angle ABF what condition should be added between the sides of parallelogram ABCD? Please fill in the condition and explain
A triangle and a parallelogram coincide, and two points a and B coincide

1. Certification:
∵AD∥BC
∴∠ADC＝∠C
∵ e is the midpoint of CD
∴CE＝DE
∵∠BEC＝∠FED
The ∧ BEC is equal to the ∧ fed (ASA)
∴BE＝EF
E is the midpoint of BF
2. Adding condition: ab = 2ad
∵ parallelogram ABCD
∴AD＝BC
∵△ BEC is equal to △ fed
∴DF＝BC
∴AF＝AD+DF＝AD+BC＝2AD
∵AB＝2AD
∴AB＝AF
∴∠F＝∠ABF

Take points E and F on the sides AB and ad of parallelogram ABCD, so that AE = AB / 3, AF = ad / 4, connect the diagonal AC of EF to g, then the value of Ag / AC is?

The intersection point of EF and AC is n, the parallel EF of DM is made through D, the triangle AFG is similar, the triangle ADM, Ag / am = AF / ad = 1 / 4, the triangle age is similar, the triangle DMC, Ag / cm = AE / DC = 1 / 3, Ag / AC = 1 / 7

Given the parallelogram ABCD, point E is a point on edge AB, AE = 3bE, point F is a point on line ad, AF = 2fd, EF intersects AC with G, the value of agcg is______ ．

Let be = a, then AE = 3A, ab = 4A. In ∵ parallelogram ABCD, ab ∥ CD, ab = CD = 4A, ∥ AEF ∥ DHF, ∥ dhae = FDAF, ∥ AF = 2fd, ∥ dhae = 12, that is, DH = 12ae = 32a, ∥ ch = 4A + 32A = 112a, ∥ ab ∥ CD, ∥ AEG ∥ CHG, ∥ agcg = aech =

As shown in the figure, the quadrilateral ABCD is the inscribed square of ⊙ o, P is the midpoint of the arc AB, and PD and ab intersect at point E, then Pede=______ ．

Connect OP, intersect AB at point F, and connect AC. according to the deduction of vertical diameter theorem, we get op ⊥ AB, AF = BF. According to the circular angle of 90 degrees, the chord is diameter, then AC is diameter. Let the side length of a square be 1, then AC = 2, and the radius of a circle be 22. According to the properties of a square, we get ∠ oaf = 45 degrees. So of = 12, PF = 2 − 12. ∵ op ∥ ad, ∥ Pede = PFAD = 2 − 12