What is the content of the "projection theorem" in mathematics?

Projection Projection is an orthographic projection. From a point to the vertical foot that passes through the vertex and is perpendicular to the base, it is called the orthographic projection of this point on this straight line. The line segment between the two endpoints of a line segment on the orthographic projection on a straight line is called the orthographic projection of the line segment on the straight line, which is the projection theorem. [Edit this paragraph] Right-angled triangle projection theorem Right-angled triangle projection theorem (also called Euclid's theorem): In a right-angled triangle, the height on the hypotenuse is the proportional median of the projection of the two right-angled sides on the hypotenuse. Each right-angled side is the median of the ratio between the projection of the right-angled side on the hypotenuse and the hypotenuse.

The formula is as shown in the figure. In Rt△ABC, ∠ABC=90°, and BD is the height on the hypotenuse AC. Then there is a projection theorem as follows:

(1) (BD) ^2;=AD·DC,

(2)(AB)^2;=AD·AC,

(3)(BC)^2;=CD·AC.

Proof: In △BAD and △BCD, ∠A+∠C=90°, ∠DBC+∠C=90°, ∴∠A=∠DBC, and ∵∠BDA=∠BDC=90° , ∴△BAD∽△CBD is similar, ∴ AD/BD=BD/CD, that is, (BD)?=AD·DC. The rest are similarly provable. (The Pythagorean theorem can also be used to prove)

Note: The Pythagorean theorem can also be proved by the above projection theorem. From 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 the Pythagorean Theorem. [Edit this paragraph] The Projection Theorem of Any Triangle The Projection Theorem of any triangle is also called the "First Cosine Theorem":

Suppose the three sides of ⊿ABC are a, b, and c, and the angles they oppose are respectively A, B, C, then there are

a＝b·cosC＋c·cosB,

b＝c·cosA＋a·cosC,

c＝a·cosB+b ·cosA.

Note: Taking "a=b·cosC＋c·cosB" as an example, the projections of b and c on a are b·cosC and c·cosB respectively, hence the name Projection Theorem.

Proof 1: Suppose the projection of point A on straight line BC is point D, then the projections of AB and AC on 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 be proved in the same way.

Proof 2: From the 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. Others can be proved in the same way.