Traditional Culture Encyclopedia - Photography major - What is the projective theorem and how to use it?
What is the projective theorem and how to use it?
Projection is orthographic projection, and the vertical foot perpendicular to the bottom from a point to a vertex is called orthographic projection of that 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. 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. The formula is shown in the figure. At 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; Equal product formula (4) abxbc = bdxac proves that in △BAD and △BCD, ∠ A+∠ C = 90, ∠ DBC+∠ C = 90, ∴ A = ∠ DBC, ∞. = A.D. DC. Everything else is similar. Note: Pythagorean theorem can also be proved by the above projective 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 Pythagorean theorem. [Edit this paragraph] The projective theorem of any triangle is also called the "first cosine theorem": let the three sides of ⊿ABC be A, B and C, and the angles they subtend are A, B and C, then there are A = B COSC+C COSB, B = C COSA+A COSC, and C = A COSC. 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, BD = CCCOSB, 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.
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