Traditional Culture Encyclopedia - Photography and portraiture - Projective Theorem in Triangle
Projective Theorem in Triangle
AC*BC=2 S ABC
CD*AB=2 S ABC
AC*BC=AB*CD
abstract
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. In the formula Rt△abC, ∠ ACB = 90, and CD is the height on the hypotenuse AB, then the projective theorem is as follows: (1) (CD) 2; =AD DB,(2)(bc)^2; =BD BA,(3)(ac)^2; =AD AB. Equal product formula (4)ACXBC=ABXCD (proof of usable area)
Projective Theorem of Folding Right Triangle
The so-called projection is the projection of light. 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:
Height theorem of right triangle
Folding proof
Solution:
In △BAD and △ACD,
∠∠Abd+∠Bad = 90, and ∠ CAD+∠ C = 90,
Schematic diagram of projective theorem
∴∠ABD=∠C,
∠∠BDA =∠BDC = 90。
∴△BAD∽△CBD
∴ AD/BD=BD/CD
Namely BD? = ad DC
The rest can be proved in the same way.
Height theorem of right triangle
Fold content
AB? =AD AC,BC? =CD CA
Two formulas have been added:
AB? +BC? =AD AC+CD AC =(AD+CD) AC=AC? (namely Pythagorean theorem).
Note: AB? It means AB quadratic.
certificate
It is known that the middle angle of triangle A is 90 degrees and the AD is high.
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。 ? The rest can also be proved.
Fold any triangle
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 you have it.
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.
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