How to solve a math plane geometry problem plane geometry ?

The simplest solution: when a, e overlap, f, b overlap, and the conditions given by the problem are satisfied, then edf= adb=45°

Extend ba to g so that ag=cf, prove dge def, and the rest is just as good as it gets. If you can ask this question, you will definitely be able to do it well with a little bit of dialing. 45°

45 degrees Rotate the triangle DFC clockwise so that the DC coincides with the DA.

Rotation is the easiest, and it can also be transformed with trigonometric identities in high school, which is the original question in the book.

The answer is: volume is equal to 1 1 1 7 = 7 (cubic centimeters).

Analysis: The volume of each small cube is 1 1 1 = 1 (cm3), and this cube figure is composed of 7 such small cubes, and its volume is 1 7 = 7 (cubic centimeters), (it can also be seen as the volume of a cube with an edge of 2 cm minus the volume of a small cube with an edge length of 1 cm).

A cube with an edge length of 2 cm digs a small cube with an edge length of 1 cm from a vertex, removes the area of three squares with a side length of 1 cm, and adds the same area of three squares with a side length of 1 cm, and its surface area remains unchanged, and its surface area can be obtained according to the surface area formula "s=6a2" of the cube.

Regular hexahedrons have the following characteristics:1. The hexahedron has 8 vertices, and each vertex is connected to three edges.

2. The regular hexahedron has 12 edges, each of which is of equal length.

3. The regular hexahedron has 6 faces, each of which has an equal area and exactly the same shape.

Fourth, the body diagonal of the hexahedron: 3a, where a is the edge length.

Take a look below and click to enlarge:

I'll offer another way of thinking, using some of the coordinates theorem and the harmonic point theorem.

Angular element Seva's theorem.

The angle sought is 30 degrees.

Let the tangent equation be y-2=k(x-2), i.e., kx-y-2k+2=0

Distance from the center of the circle (1,0) to the tangent = radius, i.e. |k-0-2k+2|/√(k²+1)=1

The solution gives k=3 4, the tangent equation is 3x-4y+2=0, and when the other tangent is k, x+c=0, |1+c|1 = 1, the solution gives c = -2, c = 0 (rounded).

The other tangent is x=2

The intersection point of the tangent and the parabola m: substituting y = 2x into 3x-4y + 2 = 0 to solve y=2 3, x=2 9, m(2 9, 2 3).

The intersection point n of the tangent and the parabola: substituting x=2 into y = 2x, the solution is y= 2, n(2, -2).

Linear mn equation: (y+2) (2 3+2) = (x-2) (2 9-2), i.e. 3x+2y-2=0

Let the straight line be y-2=k(x-2), i.e., y=kx-2k+2 substituted (x-1) 2+y 2=1

The discriminant formula is given by =0 to (-1-4k +4k) 4(1+k)(4k +4-8k)=0

i.e. 16k-12=0

The solution is k=3 4

y=(3 4)x+1 2, which is a straight line.

However, from the graph, there is also a tangent point, at (2,0), which is perpendicular to the x-axis of the line x=2 with the center of the circle, so there is also a straight line, x=2

Let h1, h2, and h3 be the points where pg1, pg2, and pg3 intersect with ab, bc, and ac, respectively, then g1g2 h1h2, g2g3, h2h3, and h2h3 are respectively. So, G1G2 is ABC, G2G3 is ABC. Again, G1G2 intersects G2G3, so face G1G2G3 face ABC

Lengthen PG1 and PG2 to AB and BC over D and E.

G1 is the center of gravity of Pab, Pg1 Dg1 2, and G2 is the center of gravity of Pbc, Pg2 eg2 2, get: Pg1 Dg1 Pg2 eg2, G1G2 de.

G1 is the center of gravity of the PB, D is the midpoint of the AB, and G2 is the center of gravity of the PBC, and E is the midpoint of the BC, de AC.

From G1G2 de, de ac, we get: G1G2 AC, G1G2 face ABC, in the same way, there is: G2G3 surface ABC, and G1G2, G2G3 are two intersecting straight lines of surface G1G2G3, surface G1G2G3 surface ABC.

So easy a The center of attention is the intersection of the midline.

Parallel lines of ab through g1, intersecting pa pb at points e ,f. Connect EG3 and extend to PC to hand over to G. Obviously, eg is also parallel to ac. Hence the proof.

Take g1g2g3 as the starting point to make perpendicular lines, and prove that the three perpendicular lines are parallel and equal.

You need the god of geometry competition.