The edge length of rectangle ABCD is ab = 2, BC = 3, point P is a point on the edge of AD, q is any point on the edge of BC, connecting AQ, DQ, passing through point P as PE parallel DQ intersecting AQ at point E, PE parallel AQ intersecting DQ at point F. let the length of AP be x, try to find the functional relationship between the area of triangle PEF and X, and find out where the point P is, and what is the maximum area of triangle PEF? When the point q is where, the perimeter of the triangle ADQ is the smallest? (the process or method to determine where q is should be given, without proof)

The edge length of rectangle ABCD is ab = 2, BC = 3, point P is a point on the edge of AD, q is any point on the edge of BC, connecting AQ, DQ, passing through point P as PE parallel DQ intersecting AQ at point E, PE parallel AQ intersecting DQ at point F. let the length of AP be x, try to find the functional relationship between the area of triangle PEF and X, and find out where the point P is, and what is the maximum area of triangle PEF? When the point q is where, the perimeter of the triangle ADQ is the smallest? (the process or method to determine where q is should be given, without proof)

The area of △ ADQ = (1 / 2) * 2 * 3 = 3 from △ ape ∽ ADQ, △ ape area / △ ADQ area = AP ^ 2 / 3 ^ 2, △ ape area = x ^ 2 / 3, similarly: △ PDF area / △ ADQ area = (3-x) ^ 2 / 3 ^ 2 △ PDF area = (1 / 3) (3-x) ^ 2, so the area of parallelogram peqf = △ ADQ area - △ ape area - △ PDF area = 3-x ^ 2 / 3

As shown in the figure, point P is a point on the edge ab of rectangle ABCD, ab = 3, BC = 4, BP = 2AP, q is a point on the edge ad, when DQ is equal to, the triangle with a, P, Q as vertex
Triangles are similar to △ PCB

According to BP = 2AP, BP = 2, AP = 1, middle angle a = angle B = 90 degrees, BC = 4, so Pb / BC = 1 / 2. When AP / AQ = 1 / 2 or AQ / AP = 1 / 2, △ PCB and △ Apq are similar, so AQ = 2 or AQ = 1 / 2, so DQ = 2 or DQ = 3,5, so when DQ = 2 or DQ = 3,5, △ PCB and △ Apq are similar

As shown in the figure, in rectangular ABCD, e is the midpoint of BC, ed intersects AC at point P, DQ ⊥ AC at point Q, ab = 2BC, AQ = QP

It is proved that: ∵ quadrilateral ABCD is a rectangle, ∵ ad = BC, DC = AB, ad ∥ BC, ∠ d = 90 °, ∵ e is the midpoint of BC, ∵ ad = BC = 2ce, ∵ ad ∥ CE, ∵ ADP ∥ CEP, ∵ adce = DPPE, ∵ ad = 2ce, ∥ DP = 2PE, that is, DP = 23de, ? CD = AB = 2BC = 22ce

As shown in the figure, in the parallelogram ABCD, m.n.p.q are AB.BC.CD . Da, and am = BN = CP = DQ, mnpq is a parallelogram

It is known that m, N, P and Q are the middle points of the edges AB, BC, CD and Da of the spatial quadrilateral ABCD respectively. It is proved that mnpq is a parallelogram

It is proved that: (1) ∵ m and N are the midpoint of AB and BC, ∵ Mn ∥ AC, Mn = AC
∵ P and Q are the midpoint of CD and DA, ∵ PQ ∥ Ca, PQ = ca
∥ Mn ∥ QP, Mn = QP, mnpq is a parallelogram
The diagonals MP and NQ of mnpq intersect and bisect each other
(2) From (1), AC ∥ Mn. Denote plane MNP (plane mnpq) as α
Otherwise, if AC &; α,
From a ∈ α, m ∈ α, B ∈ α is obtained;
From a ∈ α, Q ∈ α, we get D ∈ α, then a, B, C, D ∈α,
It is a contradiction with the known quadrilateral ABCD
And ∵ Mn &; α, ∵ AC ∥ α,
Also, AC &; α, AC ∥ α, i.e., AC ∥ plane MNP
In the same way, it can be proved that BD ‖ plane MNP

In the parallelogram ABCD, m, N, P and Q are points on AB, BC, CD and Da respectively, and am = BN = CP = DQ
So is the quadrilateral mnpa a parallelogram? Explain the reason

Yes, because the condition of parallelogram is that the opposite sides are equal, just prove that the triangle BMN is all equal to DPQ and the triangle AMQ is all equal to CPN

As shown in the figure, in the parallelogram ABCD, ab = ad, CB = CD, the points mnpq are AB, BC, CD, Da respectively, and am = BN = CP = DQ
Verification: the quadrilateral mnpq is a parallelogram

DQ = BN, ad = BC, so AQ = CN,
Angle QAM = angle NCP,
AM=CN.
So the triangle AQM is equal to the triangle CNP,
So QM = PN,
Similarly, QP = Mn
So mnpq is a parallelogram
If you draw a picture, you have to draw a picture~~~

The spatial quadrilateral ABCD and mnpq are AB.BC.CD Da, and am / MB = CN / Nb = CP / PD = AQ / QD = k prove that mnpq is coplanar and mnpq is parallelogram

It is known that the parallelogram ABCD is an inscribed quadrilateral of ⊙ o

ABCD is a quadrilateral inscribed in a circle
∴∠A+∠C=90°
ABCD is a parallelogram
∴∠A=∠C
∴∠C=90°
The parallelogram ABCD is a rectangle

It is known that, as shown in the figure, the quadrilateral ABCD is inscribed in ⊙ o, and ab ∥ CD, ad ∥ BC

It is proved that ab ∥ CDAD ∥ BC ﹥ ABCD is a parallelogram ﹥ a = ﹥ CABCD is inscribed in ⊙ o ﹥ a + ﹥ C = 180 ﹥ a = 90 ﹥ parallelogram ABCD is a rectangle