In the right angle trapezoid ABCD, AD / / BC, ∠ B = 90 °, ad = 24cm, BC = 26cm, the moving point P moves rapidly from a along the side of ad to D at the speed of 1cm per second In the right angle trapezoid ABCD, AD / / BC, ∠ B = 90 °, ad = 24cm, BC = 26cm, the moving point P starts from a and moves at the speed of 1cm / s along the ad side to D, the moving point Q starts from point C and moves at the speed of 3cm / s along the CB side to B, and P and Q start from points a and C at the same time. When one point reaches the end point, the other point also stops moving?

Let P move from a to the present point P and Q move from C to the present point Q after T seconds
∵ starting from a, the moving point P moves along the edge of ad to point d at a speed of 1cm / s
∴AP=1×t=t
∵ starting from point C, the moving point Q moves along the CB side to B at a speed of 3cm / s
∴CQ=3t
∵ in rectangular trapezoid ABCD, ad ∥ BC, ∠ B = 90 degrees, ad = 24cm, BC = 26cm, the vertical line through P is BC, the vertical foot is f, the vertical line through D is BC, the vertical foot is g, the vertical line through q is ad, the vertical foot is e, then CG = BC-AD = 26cm-24cm = 2cm. If and only if QF = CG = 2cm, the quadrilateral PQCD is isosceles trapezoid
In this case, EP = QF = 2cm
∵AD =AP+ PD=24cm,PD= CQ－CG－FQ=3t－2－2=3t－4
∴t+3t－4=24
T = 7 seconds
When t is 7 seconds, the quadrilateral PQCD is isosceles trapezoid

In rectangular trapezoid ABCD, ad ∥ BC, ab = 8cm, ad = 24cm, BC = 26cm. The moving point P moves from a to D at a speed of 1cm / s, and the moving point Q moves from C to B at a speed of 3cm / s. let t be the moving time
Q: can a quadrilateral abqp be a square at a certain time?
I don't think it's OK. Because AB = 8, if it's a square, then AP is equal to 8, that is to say, 8 seconds. But in 8 seconds, Q moves 3 × 8 = 24s, 26-24 = 2cm, BQ = 2cm, so it's not a square
Will it become a square

I agree with your explanation. It won't be a square

As shown in the figure, in the rectangular trapezoid ABCD, ab ‖ CD, ad ⊥ CD, ab = 1cm, ad = 2cm, CD = 4cm, then BC=______ cm．

As shown in the figure, if we make be ⊥ CD through point B, then the quadrilateral abed is a rectangle, ≁ ad = be = 2cm, de = AB = 1cm ≁ CE = cd-de = 4-1 = 3cm ≁ BC = be2 + CE2 = 13cm

In the quadrilateral ABCD, ∠ a = 60, ab ⊥ BC, ad ⊥ DC, ab = 20, CD = 10, find the length of AB and CD (the answer retains the root)
(2) Finding the area of quadrilateral ABCD
Wrong, the first to find the length of AD and BC

Extend ad, BC to point e
∵AB⊥BC
∴∠B=90°
∵∠A=60°
∴∠E=30°
∴AE=2AB=2×20=40
(in a right triangle, the right side of a 30 ° angle equals half of the hypotenuse)
BE=√﹙AE²-AB²)=20√3
∵AD⊥DC
∴∠CDE=90°
∵∠E=30°
∴CE=2CD=20
DE=√﹙CE²-CD²)=10√3
∴AD=AE-DE=40-10√3
BC=BE-CE=20√3-20
(2) Area of quadrilateral ABCD
=Area of triangle Abe area of triangle CDE
=½×20×20√3-½×10×10√3
=150√3

It is known that in the trapezoidal ABCD, ad ‖ BC, e and F are the midpoint of AB and CD respectively, connecting EF. Please use the method of proving the triangle median theorem. Gvrdk

Because ad ∥ BC, GA: ab = Gd: DC, because e is the midpoint of AB, f is the midpoint of CD, so Ge: EB = (GA + 1 / 2Ab): (1 / 2Ab) = 2ga: ab + 1 GF: FC = (GD + 1 / 2CD): (1 / 2CD) = 2Gd: CD + 1 = 2ga: ab + 1 = Ge: EB, so EF ∥ BC. Because EF ∥ BC, so HF ∥ BC, eh ∥ ad, and H is the midpoint of BD. according to the triangle median theorem, eh = 1 / 2ad, HF = 1 / 2BC, so EF = 1 / 2 (AD + BC)

Given the rectangular trapezoid ABCD, ad parallel BC e is the hypotenuse, and the midpoint of CD is made EF ⊥ AB, it is proved that EF is the median line of rectangular trapezoid ABCD

Certification:
Connect DF and extend, and cross the extension line of CB to point G
∵EF‖BC
Then △ def ∽ DGC
∴DF/DG=DE/DC
∵DE =CE
∴DF=FG
∵AD‖BG
Then △ ADF ≌ △ BGF
∴AF=BF
The median line of ABCD is Fe

As shown in the figure, in the isosceles trapezoid ABCD, ab ‖ DC, e is a point on the DC extension line, be = BC, try to explain the relationship between ∠ A and ∠ E

Reasons: ∵ trapezoid ABCD is isosceles trapezoid ∵ a = ∵ ABC (2 points) ∵ be = BC ∵ e = ∵ BCE (4 points) and ab ∥ DC ∵ ABC = ∵ BCE (6 points) ∵ a = ∵ e (8 points)

As shown in the figure, in isosceles trapezoid ABCD, ad is parallel to BC, AC is vertical to BD, and BC trace point E is extended to make CE = ad. try to judge the shape of triangle BDE

Isosceles right triangle
Because in quadrilateral adce
Ad parallel CE and ad equal CE
So the quadrilateral adce is a parallelogram
So AC is parallel and equal to de
So the angle BOD equals the angle BDE equals 90 degrees
And because BD equals AC equals De
So the triangle BDE is a right triangle

As shown in the figure, it is known that the lengths of AB and BC on both sides of rectangle ABCD are 6 and 10 respectively. E is a point on ab. fold up the triangle EBC along CE. If point B just falls at f on the edge of AD, find the position of point E on ab

"B just falls at f on the edge of ad" indicates two points: be = ef (set as x), then AE = 6-x
CF = BC = 10, because DC = 6, then DF = 8, AF = 2,
In triangle AEF, AE'2 + AF'2 = EF'2
That is: X'2 = (6-x)'2 + 2'2
X = 10 / 3

In the isosceles triangle ABCD, ad is parallel to BC, point E is a point on the extension line of AD, de equals BC, prove that angle e equals angle DBC

Your question is wrong. It should be in the quadrilateral ABCD BC.AD (DE) parallel BC
Then BCDE. Forms a parallelogram with equal diagonally