In the isosceles trapezoid ABCD, the upper sole ad = 2, the lower sole BC = 8, M is the midpoint of the waist AB, if MD is perpendicular to CD Don't copy other people's answers, Besides, it's troublesome: For example: DQ over D ⊥ BC over Q Make point n in CD, connect Mn, and cross DQ to s Mn is a trapezoidal ABCD median line ∴MN=5,MN‖BC The MS is a trapezoidal abqd median line Ms = 7 / 2? How did you jump to this step? Finding the area of trapezoid

In the isosceles trapezoid ABCD, the upper sole ad = 2, the lower sole BC = 8, M is the midpoint of the waist AB, if MD is perpendicular to CD Don't copy other people's answers, Besides, it's troublesome: For example: DQ over D ⊥ BC over Q Make point n in CD, connect Mn, and cross DQ to s Mn is a trapezoidal ABCD median line ∴MN=5,MN‖BC The MS is a trapezoidal abqd median line Ms = 7 / 2? How did you jump to this step? Finding the area of trapezoid

Because ad = 2, BC = 8,
So QC = (BC-AD) / 2 = (8-2) / 2 = 3
Because m is the midpoint of waist AB and N is the midpoint of waist CD
So Mn / / BC, Sn / / QC
So Sn = QC / 2 = 3 / 2
Because Mn = (AD + BC) / 2 = (2 + 8) / 2 = 5
So Ms = mn-sn = 5-3 / 2 = 7 / 2

As shown in the figure, we know that in the rectangular trapezoid ABCD, ad ∥ BC, ab ⊥ BC, ad = 2, ab = 8, CD = 10;
In the process of motion, is there such a t that the triangle with P, D and Q as the vertex is exactly the isosceles triangle with DQ as the waist? If so, all the values of t that meet the conditions are requested; if not, please explain the reason

(1) Make de vertical BC through point D, and the perpendicular foot is e. ∵ ad ∥ BC ab ⊥ BC ∪ a = 90 ° ∨ CE ⊥ BC (done) ∪ bed = 90 ° ∪ quadrilateral abed is rectangle (quadrilateral with three right angles is rectangle) ∪ AB = de = 8 ∫ CD = 10. According to Pythagorean theorem, we get: CE = √ DC's Square - De's Square = 6 ∪

As shown in the figure, in the parallelogram ABCD, ab = 2BC, M is the midpoint of AB, the proof is cm ⊥ DM

As shown in the figure, extend DM, CB, two lines intersect at the point E, ∵ ad ∥ BC, ∵ dam = ∵ EBM. ∵ am = BM, ∵ amd = ≌ BME, ≌ ADM ≌ BEM, ≌ DM = em, ad = be. ∵ AB = CD, ab = 2CB, ∵ CD = 2CB, ≁ CD = CE, ≁ cm ⊥ DM

In parallelogram ABCD, diagonal lines AC and BD intersect at point O, point P is a point outside the quadrilateral, and PA ⊥ PC, Pb ⊥ PD, the perpendicular foot is p

It is proved that the connection OP, ∵ PA ⊥ PC, Pb ⊥ PD, ∵ APC and ⊥ BPD are right triangles, ∵ quadrilateral ABCD is parallelogram, ∵ Ao = co = 12ac, Bo = do = 12dB, ∵ in right angle ⊥ APC, OP is hypotenuse midline, ∵ OP = 12ac, ∵ in right angle ⊥ BPD, OP is hypotenuse midline, ∵ OP = 12bd, ∵