Classical problems about similar projective theorem of right triangle in junior middle school (with solution)

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outstanding

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.

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Right triangle 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, ∠ BAC = 90, and AD is the height on the hypotenuse BC, the projective theorem is as follows:

( 1)(AD)^2=BD？ DC，

(2)(AB)^2=BD？ BC,

(3)(AC)^2=CD？ In 200 BC.

It is proved that in △BAD and △ACD, ∠ B+∠ C = 90, ∠ DAC+∠ C = 90, ∠ BDA = DC. The rest are almost the same.

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

(AB)^2+(AC)^2=BD？ BC+CD？ BC =(BD+CD)？ BC=(BC)^2，

That is, (AB) 2+(AC) 2 = (BC) 2.

This is the conclusion of Pythagorean theorem.

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Projective Theorem of Arbitrary Triangle

The projective theorem of arbitrary triangle 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: Use "A = B? cosC+c？ CosB ",for example, the projection of B and C on A is B? cosC、c？ CosB, hence the name 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。 The rest can also be proved.