Let the vertex a of the square ABCD be the plane ABCD, PA = AB = a, and find the dihedral angle between the plane PAB and the plane PCD

Let the vertex a of the square ABCD be the plane ABCD, PA = AB = a, and find the dihedral angle between the plane PAB and the plane PCD

As shown in the figure, consider the third plane ABCD which intersects with plane PAB and plane PCD at the same time. The intersection lines are AB and CD, and ab ‖ CD, then the edge of dihedral angle formed by plane PAB and plane PCD must be parallel to AB and CD. In plane PAB, PQ ‖ AB is made through point P, then PQ is the edge of dihedral angle formed by plane PAB and plane PCD

The dihedral angle formed by the plane ABP and the plane CDP is obtained when the line AP is perpendicular to the cotton grower ABCD through the fixed point a of the square ABCD and AP = ab

45 degrees!
Imagine!

As shown in the figure, it is known that the side length of a square ABCD is a. now, from its four vertices a, B, C and D, the equal segments AP and BQ are intercepted to the direction of points B, C, D and a respectively,

Let as = Dr = CQ = BP = X,
Then AP = BQ = CR = DS = A-X,
From the meaning of the title, we can get: Δ APS ≌ Δ DSR ≌ Δ CRQ ≌ Δ BQP,
S quadrilateral PQRS = s Square abcd-4s Δ APS
=a^2-4*1/2X(a-X)
=a^2-2a+2X^2
=2(X-a/2)^2+1/2a^2
When x = A / 2, smin = 1 / 2A^

Let the vertex a of the square ABCD be the plane ABCD, PA = AB = a, and find the dihedral angle between the plane PAB and the plane PCD

As shown in the figure, consider the third plane ABCD which intersects with plane PAB and plane PCD at the same time. The intersection line is ab and CD, and ab ∥ CD, then the edge of dihedral angle formed by plane PAB and plane PCD must be parallel to AB and CD. In plane PAB, pass through point P to make PQ ∥ AB, then PQ is the edge of dihedral angle formed by plane PAB and plane PCD, and then it can be proved that PA ⊥ PQ, PD ⊥ PQ, ∠ APD are the angle, which can be solved in RT △ apd It is concluded that ∠ APD = 45 °

The vertex a of the square ABCD is PA, perpendicular to the plane ABCD is be, perpendicular to the PC, the perpendicular foot is e, connecting De to find ∠ bed, which is the plane angle of the dihedral angle b-pc-d
I can't prove be vertical PC, de vertical PC,

The drawing process can be regarded as a known condition without proof
Pb = Pb from △ PAB ≌ △ pad,
If △ PBC ≌ △ PDC, then ∠ BPC = ∠ DPC,
From △ PBE ≌ △ PDE, ∠ ped = ∠ PEB = 90 °

Given that ABCD is a rectangle, ab = 3, ad = 4, PA ⊥ surface ABCD, PA = 2, q is the midpoint of PA, find the distance from Q to BD

Q is the midpoint of PA, QA = 1, right triangle ABQ known AQ = 1, ab = 3, get BQ, similarly get DQ, known ad, AB, get BD, triangle bdq known, three sides seek height, don't I say it again?

Find a point O in the quadrilateral ABCD to minimize the sum of the distances from it to the four vertices of the quadrilateral, and please give your reasons. What is the conclusion of this problem? Give an example to illustrate its application in real life

AC, BD intersection, according to: between two points, the shortest line segment!

In the quadrilateral ABCD, P is any point, AP ^ 2 + CP ^ 2 = BP ^ 2 + DP ^ 2, finding the quadrilateral must be

The answer you want is: Idea: first take a special point to push out a quadrilateral as a rectangle, and then verify that for a rectangle, any point P in the plane satisfies AP ^ 2 + CP ^ 2 = BP ^ 2 + DP ^ 2. You may as well take P as the midpoint of ab, then PC = PD can be obtained by AP ^ 2 + CP ^ 2 = BP ^ 2 + DP ^ 2. If the midpoint of CD is Q, then PQ is perpendicular to CD, and PQ is also perpendicular to ab

Square ABCD side length is 8, AP = CP = 5, find BP
Square ABCD side length is 8, AP = CP = 5, P is a point in the square, find the BP length
AP = CP = radical 5

How can it be ap = CP = √ 5? Because the side length is 8, then AC = 8 √ 2, P is a point in the square, then AP + CP ≥ 8 √ 2, 8 √ 2 > 10, so AP = CP = 5 is not right

The square ABCD is inscribed with O, and the point P is a point on the inferior arc ad. it connects AP, BP, CP and calculates (AP + CP) / BP without trigonometric function

Root 3 / 3 + 1