It is proved that vertical lines, parallel lines, vertical lines and planes, vertical lines and planes, vertical planes and parallel planes are common problems in high school solid geometry, and they are interrelated and transformed with each other. At the same time, we need to take appropriate actions to achieve our goals. By integrating the knowledge points learned before and after, the task of proof is completed by various methods to achieve the goal of internalization of analogy. This paper introduces a problem of proving the vertical line by various methods, hoping to get through all kinds of proof ideas, consolidate the basic skills of learning and bring some enlightenment and help to everyone.

First of all, the question is raised:

As shown in the figure, in the quadrangular cone P-ABCD, the bottom ABCD is a right-angled trapezoid, with ad∨BC, ∠ Bad = 90, PA⊥ bottom ABCD, PA=AD=AB=2BC, and M is the midpoint of PG.

(1) verification: Pb ⊥ DM;

(2) Find the cosine of the angle formed by AC and PD.

The following focuses on the various proofs of (1) quiz, omitting the answers of (2) quiz.

Proof 1: coordinate method.

As shown in figure 1, take point A as the coordinate origin, and establish the spatial rectangular coordinate system A-xyz as shown in the figure. Let PA=AD=AB=2BC=4, then A (0 0,0,0), B (4 4,0,0), C (4 4,2,0) and D (0 0,4,0).

Note: This method uses the "coordinate method" to calculate the zero product of vector PB and vector DM, which proves that PB⊥DM needs to establish a suitable spatial rectangular coordinate system and express the coordinates of related points one by one. It is noted that the multiple relationship between line segments is known, and the length of long line segments is set to a constant of 4, which is convenient to express all vector coordinates by integers, thus simplifying the calculation. This method requires that the relevant points must be accurately expressed and the idea of combining numbers and shapes should be applied.

Prove 2: Apply the normal vector of a straight line parallel to the plane of another straight line.

As shown in Figure 2, the spatial rectangular coordinate system A-xyz as shown in the figure is established with point A as the coordinate origin. Let PA=AD=AB=2BC=4, connect AM, then A (0,0,0), B (4,0,0), C (4,2,0), D (0,4,0).

Comment: It is not difficult for us to find PAB of AD⊥ aircraft, get PB⊥AD, prove this method PB⊥DM, from which we can infer ADM of PB⊥ aircraft. So the problem can be transformed into proving that the normal vectors of PB and ADM are parallel vectors.

Proof 3: Basis Vector Method.

The above picture continues:

The above picture continues:

Comment: This method uses "basis vector method" to prove PM⊥DM. Firstly, vector AB, vector AD and vector AD are taken as a set of basis vectors, and the length of the longer line segment in the topic is set to 2, and then it can be proved by vector operation. This method uses the equivalence of the zero product of the number of vertical vectors many times, and proves that the process is straightforward, and the goal can be achieved only by patient operation.