What is the projective theorem in junior high school mathematics? How to prove it?

The so-called projection is the orthographic projection. Right triangle projection theorem (also called Euclid theorem): In a right triangle, the square of the height on the hypotenuse is the proportional average of the projection 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). Equal product formula (4)AB×BC=AC×BD (can be proved by "area method") Proof sketch of right triangle projection theorem (geometric sketchpad)

(mainly calculated from the similarity ratio of triangles) 1. In △BAD and △BCD, ∵ Abd+∠ CBD = 90, while ∠ CBD+∠ C = 90, ∴∠Abd =∞. The rest can be proved in the same way: Pythagorean theorem can also be proved by the above projective theorem. The projective theorem is as follows: AB 2 = AD AC, BC 2 = CD Ca, which is the sum of AB 2+BC 2 = AD AC+CD AC = (AD+CD) AC = AC 2. That is, AB 2+BC 2 = AC 2 (tick second, the angle A in the known triangle is A=90 degrees, and AD is high. The projection ∵ AD 2 = AB 2-BD 2 = AC 2-CD 2 is proved by Pythagoras, ∴ 2ad2 = ab+AC-BD-CD = BC-BD-CD = (BD+CD)-(BD+CD) = 2bd× CD. Therefore, AD 2 = BD× CD. Using this conclusion, we can get: AB = BD+AD = BD+BD× CD = AC = CD+AD = CD+BD× CD = CD (BD+CD) = CD× CB. The projective theorem of any triangle is also called the "first cosine theorem": the three sides of △ABC are A, B and C, and the angles they face are A, B and C, so there is A = B Cosc+C Cosb. Note: Take "A = B COSC+C COSB" as an example, and the projections of B and C on A are B COSC and c cosc respectively. 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, BD=c cosB, CD = b cosc, ∴ a = BD+CD = b cosc+c cosb. The rest can also be proved.

Prove 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.