In the pyramid p-abcd, the quadrilateral ABCD is rhombic, Pb = PD, and E and F are the midpoint of BC and CD respectively

In the pyramid p-abcd, the quadrilateral ABCD is rhombic, Pb = PD, and E and F are the midpoint of BC and CD respectively

It is proved that: (1) ∵ E and F are the midpoint of BC and CD respectively, ∵ EF / / BD ∵ EF is parallel to a straight line on the plane PBD ∵ EF / / plane PBD (2) ∵ ABCD is a diamond, so BD ⊥ AC, ∵ EF / / BD, ∵ EF ⊥ AC let AC and BD intersect at O, then Bo = OD, connect Po ∵ Pb = PD, ≁ Po ⊥ BD, so EF ⊥ Po ≁ EF ⊥ plane PAC from

In rectangular ABCD, M is the midpoint of ad side, P is the point above BC, PE ⊥ MC, PF ⊥ MB, when AB and BC satisfy the condition______ The quadrilateral PEMF is rectangular

When AB = 12bc, the quadrilateral PEMF is a rectangle. ∵ in rectangle ABCD, M is the midpoint of ad side, ab = 12bc, ∵ AB = DC = am = MD, ∵ a = ∵ d = 90 °, ∵ ABM = ∵ MCD = 45 °, ∵ BMC = 90 ° and ∵ PE ⊥ MC, PF ⊥ MB, ∵ PFM = ∵ PEMF = 90 °, ∵ quadrilateral PEMF is a rectangle

Known: as shown in the figure, P is a point on the edge of CD of rectangular ABCD, PE ⊥ AC in E, PF ⊥ BD in F, AC = 15, BC = 8, find PE + PF

In RT △ PCE, sin ∠ DCA = sin α = PEPC, ｜ PE = pcsin α, in RT △ PDF, sin ∠ BDC = sin α = pfdp, ｜ pf = pdsin α, ｜ PE + pf = pcsin α + pdsin α = cdsin α, ∵ in RT △ BCD

As shown in the figure, e is the midpoint of edge BC of rectangular ABCD, P is a moving point on edge ad, PF ⊥ AE, pH ⊥ De, and the perpendicular feet are f and h respectively
(1) When the length and width of rectangle ABCD satisfy what conditions, is the quadrangle phef a rectangle? (2) in (1), when the moving point P moves to what position, the rectangular phef becomes a square? Why?

(1) It is proved that: ∵ quadrilateral ABCD is a rectangle, ∵ ad = BC, ab = CD; ∵ e is the midpoint of BC, ∵ AB = be = EC = CD; then ∵ Abe, ∵ DCE are isosceles RT △; ∵ AEB = ∵ Dec = 45 °; ∵ AED = 90 °; in quadrilateral pfeh, ∵ PFE = ∵ FEH = ∵ EHP = 90 °, so quadrilateral pfeh is a moment

As shown in the figure, the plane of ABCD and the plane of positive △ pad are perpendicular to each other. M and Q are the midpoint of PC and ad respectively. (1) verify the MBD of PA ‖ plane; (2) calculate the distance from a to PBD of plane

(1) It is proved that if AC intersects BD with O and Mo, ABCD is a square, so o is the midpoint of AC and M is the midpoint of PC, so Mo ⊄ PA, PA ⊄ plane MBD, Mo ⊂ plane MBD, PA ⊂ plane MBD; (2) if QE ⊥ BD is used to connect PE, then the plane of square ABCD is perpendicular to the plane of positive △ pad

As shown in the figure, the plane of square ABCD with side length of 4 is perpendicular to the plane of triangular pad, and M q is the midpoint of PC and ad respectively
Question: is there a point n on the line AB such that plane PCN ⊥ plane pqb? If so, point out the position of point n and prove your theory; if not, explain the reason

According to the median line theorem, PA ∥ Mo can be obtained, and then it can be proved by the judgment theorem of parallel lines and planes
Connect AC to BD, then connect mo
∴PA∥MO
PA &; planar MBD, MO &; planar MBD
Plane MBD

As shown in the figure, in the pyramid p-abcd, plane pad ⊥ plane ABCD, ab = ad, ∠ bad = 60 °, e and F are the midpoint of AP and ad respectively. The following are proved: (1) straight line EF ∥ plane PCD; (2) plane bef ⊥ plane pad

It is proved that: (1) in △ pad, because e and F are the midpoint of AP and ad respectively, EF ‖ PD. And because EF is not in plane PCD, PD ⊂ plane PCD, so straight line EF ‖ plane PCD. (2) connect BD. because AB = ad, ∠ bad = 60 °, so △ abd is an equilateral triangle

Given the plane of the PA vertical parallelogram ABCD, if PC ⊥ BD, the parallelogram ABCD must be______ ．

According to the meaning of the question, draw a figure as shown in the figure, ∩ PC = P. ∩ BD ⊥ PAC ⊂ AC ⊂ PAC ⊂ AC ⊥ BD ⊂ ABCD, PC ⊂ ABCD, PA ∩ PC = P. ∩ BD ⊥ PAC ⊂ AC ⊂ BD ⊥ ABCD is parallelogram ⊂ parallelogram ABCD must be diamond

As shown in the figure, in the pyramid p-abcd, PA ⊥ has the bottom surface ABCD, and all sides of the bottom surface are equal. M is a moving point on PC, when m satisfies______ If you want to fill in a condition that you think is correct

According to the theorem, BD ⊥ PC.. When DM ⊥ PC (or BM ⊥ PC), there is PC ⊥ plane MBD, while PC ⊂ plane PCD, ⊥ plane MBD ⊥ plane PCD. Therefore, DM ⊥ PC (or BM ⊥ PC, etc.) is selected

As shown in the figure, in the parallelogram, ∠ ABC = 75 °. AF ⊥ BC is in F, AF intersects BD in E, if de = 2Ab, ∠ AED=______ °．

ABCD is a parallelquadrilateral, and its \\\\\\\\abcdis a parallelogram, \ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ ° and ∠ ADO= Therefore, the answer is 65 degrees