Please tell me the specific content and formula of the projective theorem (junior two mathematics).

The so-called projection is the orthographic projection. Projection theorem of right triangle (also called Euclid theorem): In a right triangle, the height on the hypotenuse is the proportional average of the projections of two right angles on the hypotenuse. Each right-angled edge is the median of the projection of this right-angled edge on the hypotenuse and the proportion of the hypotenuse.

Formula: As shown in the figure, in Rt△ABC, ∠ ABC = 90, and BD is the height on the hypotenuse AC, then the projective theorem is as follows:

(1) (BD) 2 = AD DC, (2) (AB) 2 = AD AC, (3) (BC) 2 = CD California).

Equal product formula (4)AB×BC=AC×BD (can be proved by "area method")

Proof of projective theorem of right triangle

[Sketch of Projection Theorem (Geometric Sketchpad)]

Schematic diagram of projective theorem (geometric sketchpad)

(mainly calculated from the similarity ratio of triangles) 1,

In △BAD and △BCD, ∫ Abd+∠ CBD = 90, and ∠ CBD+∠ C = 90,

∴∠ABD=∠C，

∠∠BDA =∠BDC = 90。

∴△BAD∽△CBD

∴ AD/BD=BD/CD

That is BD 2 = ad DC. The rest can be proved in the same way.

Note: Pythagorean theorem can also be proved by the above projective theorem.

The projective theorem is as follows:

California AB 2 = AD AC, BC 2 = CD.

Two formulas have been added:

ab^2+bc^2=ad AC+CD AC =(ad+CD)ac=ac^2。

That is, AB 2+BC 2 = AC 2 (Pythagorean theorem conclusion).

Second,

It is known that the middle angle of triangle A is 90 degrees and the AD is high.

Pythagoras proof projection

∵AD^2=AB^2-BD^2=AC^2-CD^2，

∴2ad=ab+ac-bd-cd=bc-bd-cd=(bd+cd)-(bd+cd)=2bd×cd.

Therefore, AD=BD×CD.

Using this conclusion, we can get: AB = BD+AD = BD+BD× CD = BD× (BD+CD) = BD× BC, AC = CD+AD = CD+BD× CD = CD (BD+CD) = CD× CB.

To sum up, the projective theorem is obtained. It can also be proved by the knowledge of triangle area.

Edit the projective theorem of arbitrary triangle in this section.

The projective theorem of arbitrary triangle is also called "the first cosine theorem";

△ABC's three sides are A, B and C, and the angles they face are A, B and C respectively, so there is.

a=b cosC+c cosB，

b=c cosA+a cosC，

c=a cosB+b cosA .

Note: Take "A = B COSC+C COSB" as an example. The projections of b and c on a are B COSC and C COSB, respectively, so there is a projective theorem.

Proof 1: Let the projection of point A on the straight line BC be point D, then the projections of AB and AC on the straight line BC are BD and CD respectively, and

BD=c cosB，CD = B COSC，∴ A = BD+CD = B COSC+C COSB。 The rest can also be proved.

Prove 2: From sine theorem, we can get: b=asinB/sinA, c = asinc/sina = asin (a+b)/sina = a (sinacosb+Cosasinb)/sina.

= acosb+(asinb/Sina)COSA = a COSB+b COSA。 It can also prove others.

Edit this projective theorem-area projective theorem.

Area projection theorem: "The projection area of a plane figure is equal to the area s of the projected figure multiplied by the cosine of the included angle between the plane where the figure is located and the projection plane."

COSθ=S projective /S primitive

(The area of a planar polygon and its projection are S primitive and S projection respectively, and the sharp dihedral angle formed by its plane is θ)

Proof idea: Because the projection is to scale the length of the original figure (the height in the triangle) and the width is unchanged, and because the area ratio of the plane polygon = the square ratio of the side length. So it is the ratio of the length of the figure (called the height in the triangle). Then this ratio should be the cosine of the angle formed by the plane. Make a right-angled triangle on two planes so that the hypotenuse and right-angled edge are perpendicular to the edge (that is, the intersection of the plane where the original polygon is located and the projection plane), then the hypotenuse and the other right-angled edge of the triangle are the length ratio of its polygon, that is, the area ratio of the plane polygon, and substitute this ratio into the plane triangle for operation.