In the quadrilateral ABCD, BD bisects ∠ ABC, AC ⊥ BD at R, PQ intersects BC and ad at points Q and P respectively, and ∠ bad = ∠ BQP

In the quadrilateral ABCD, BD bisects ∠ ABC, AC ⊥ BD at R, PQ intersects BC and ad at points Q and P respectively, and ∠ bad = ∠ BQP

In △ ABR and △ CBD, ab = BC, ab = BC, ab = BD, ab = BD, ab = BD, ab = BC, ab = BD, ab = BD, ab = BD, ab = BD, ab = BD, ab = SAS, ab = BD, ab = BD, ab = BD, ab = SAS, ab = BC, ab = BC, ab = BC, ab = BC, ab = BD, ab = BD, ab = BD, ab = SAS, ab = BC, ab = BC, ab = BC, ab = BC, ab = BC, ab = BC, ab = BD, ab = BD, ab = BD, ab = BD, ab = SAS, ab = AB = BC, ab = BC, ab = BC, ab = BC, ab = BC, ab = BC, ab = BC, ab = BC, ab = BD, ab = BD, ab = BD, ab = BD, ab = BD, ab = BD, ab = BD, ab = SAS

As shown in the figure, P is a point outside ABCD, Pb = 10 cm, the area of triangle APB is 40 square cm, and the area of triangle CPB is 30 square cm
What's the area of a square?

Let the height of AB in APB be p, the height of BC in CPB be q, and the side length of a square be a
The area of triangle APB = AB * P / 2 = 40
The area of triangle CPB = BC * q / 2 = 30
From Pythagorean theorem, P ^ 2 + Q ^ 2 = 100
The equation (80 / a) ^ 2 + (60 / a) ^ 2 = 100 can be obtained
The solution is a = 10
The area is 100

As shown in the figure, P is a point outside ABCD, Pb = 10 cm, triangle APB area is 40 square cm, triangle CPB area is 20 square cm
The area of square ABCD is

120

As shown in the figure, let m and n be the midpoint of AD and CB at the waist of rectangular trapezoid ABCD respectively, AB on De is at point E, and △ ade is folded along De, and m and N coincide exactly, then AE: be is equal to ()
A. 2：1B. 1：2C. 3：2D. 2：3

Let de and Mn intersect at point F, ∵ m and n be the midpoint of AD and CB respectively, ∵ Mn ∥ AB, and ∵ m be the midpoint of AD, ∵ MF = 12ae, and ∵ m and N coincide, ∵ NF = be, MF = NF, ∵ AE: be = 2mf: NF = 2:1, so a

As shown in the figure, let m and n be the two waist ad of right angle trapezoid ABCD and the midpoint of CB respectively, and let de ⊥ AB be in E. flip △ ade along De, and let m and N coincide exactly, then AE: be is?

If △ ade is flipped along De, m and N coincide exactly. Therefore, suppose that connecting Mn and de intersect at point G, then mg = ng, and the right part is a rectangle, so mg = ng = be

In the known trapezoid ABCD, ab ‖ CD (1) is shown in the figure, point m is the midpoint of waist BC, and am = DM
In the known trapezoidal ABCD, ab ‖ CD
(1) As shown in the figure, point m is the midpoint of waist BC, and am = DM, proving that trapezoid ABCD is a right angle trapezoid
(2) As shown in the figure, the point n is the midpoint of the bottom CD, and an = BN, proving that the trapezoid ABCD is isosceles trapezoid
[= = sorry, I can't upload the picture, but I should be able to draw it]

1. Take the midpoint n of AD and connect Mn, then Mn is the median line of trapezoid, ∧ Mn ∧ AB, ∧ am = DM, ∧ Mn ⊥ ad (isosceles triangle three lines in one), ∧ ab ⊥ ad, that is, trapezoid ABCD is right trapezoid. 2, ∧ ab ∧ CD, ∧ NAB = ∧ and, ∧ NBA = ∧ BNC, ∧ an = BN, ∧ nab = ∧ NBA, ∧ and = ∧ BNC, ∧ DN = CN

As shown in the figure, in the trapezoid ABCD, ad ‖ BC, points E and F are the midpoint of AB and CD respectively. Like EF, we call the segment connecting the midpoint of the two waists of the trapezoid the median line of the trapezoid

Extend CD, then BC, and cross point M
Triangle ADF congruence and triangle MCF (angle, side, angle),
So ad BC = BM,
At this time, the trapezoidal median line becomes the median line of the triangular AMB,
So the median line of triangle AMB (EF) = 1 / 2bm = 1 / 2, so the upper and lower base (AD BC)

If there is a right angled trapezoidal part ABCD, AD / / BC, the length of diagonal waist DC is 10cm, the angle d = 120 degrees, ad = half BC, then how much cm is ab equal to?

The perpendicular of BC through D intersects BC at h

As shown in the figure, in ladder ABCD, ab ‖ DC, ∠ B = 90 °, e is the point on BC, and AE ⊥ ed. if BC = 12, DC = 7, be: EC = 1:2, find the length of ab

∫ ab ∥ DC, and ∠ B = 90 °, ∧ AEB + ∠ BAE = 90 ° and ∠ C = 90 degree. (1 point) ∧ AEB + ∠ CED = 90 degree. So ∠ BAE = ∠ CED. (2 points) ∧ EAB ∧ Dec. (3 points) ABEC = becd. (3 points) be: EC = 1:2, BC = 12 and DC = 7, so ab8 = 47. (4 points) AB = 327. (5 points)

As shown in the figure, in rectangle ABCD, de ⊥ AC is equal to e, AE: EC = 3:1, if DC = 6, find the length of AC

Let AE = 3k, EC = k, then AC = AE + EC = 3K + k = 4K, ∵ de ⊥ AC, ∵ CDE + ∠ ACD = 90 degree, ∵ DAC + ∠ ACD = 90 degree, ∵ DAC = ∠ CDE, and ∵ ADC = ∠ Dec = 90 degree, ∵ ACD ∽ DCE, ∵ CDAC = ECCD, that is, 64K = K6, k = 3, ∵ AC = 4 × 3 = 12