Projective projection is orthographic projection, and the vertical foot perpendicular to the bottom from a point to a vertex is called orthographic projection of the point on this straight line. The line segment between the orthogonal projections of two endpoints of a line segment on a straight line is called the orthogonal projection of this line segment on this straight line, that is, the projection theorem. [Edit this paragraph] Right triangle projection theorem Right triangle projection theorem (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.

The formula is shown in the figure. At Rt△ABC, ∠ ABC = 90, and BD is the height on the hypotenuse AC, then there is a projective theorem as follows:

( 1)(bd)^2； = ad DC,

(2)(ab)^2； =AD AC，

(3)(bc)^2； =CD AC .

It is proved that in △BAD and △BCD, ∠ A+∠ C = 90, ∠ DBC+∠ C = 90, ∴∠A=∠DBC and ∠ BDA. = A.D. DC. Everything else is similar. (It can also be proved by Pythagorean theorem)

Note: Pythagorean theorem can also be proved by the above projective theorem. According to formula (2)+(3):

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

That is (ab) 2; +(bc)^2； =(ac)^2； .

This is the conclusion of Pythagorean theorem. [Edit this paragraph] Any triangle projective theorem Any triangle projective theorem is also called "the first cosine theorem";

Suppose that the three sides of ⊿ABC are A, B and C, and the angles they face are A, B and C respectively, then there are

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.