Proof of Pythagorean Theorem 16

Pythagorean Theorem: In a right triangle, the sum of squares of two right-angled sides is equal to the square of hypotenuse, that is, in a triangle where A and B are right-angled sides and C is hypotenuse, there is a 2+B 2 = C 2.

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Evidence 1 (Zou's proof):

Make four congruent triangles, with A and B as right angles and C as hypotenuse. Put them together as shown in the figure below to make A, E and B collinear, B, F and C collinear, and C, G and D collinear.

∵Rt△HAE≌Rt△EBF

∴∠AHE=∠BEF

∠∠AHE+∠AEH = 90

∴∠BEF+∠AEH=90

∵A, E and B are collinear.

∴∠ Hef = 90, and the quadrilateral EFGH is a square.

Because the four right-angled triangles in the above picture are congruent, it is easy to conclude that the quadrilateral ABCD is a square.

∴ area of square ABCD = area of four right triangles+area of square EFGH.

∴(a+b)^2=4？ ( 1/2)? Ab+c 2, finishing a 2+b 2 = c 2.

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Proof method 2 (textbook proof):

As shown in the above figure, the areas of two squares with side length a+b are equal.

So a 2+b 2+4? ( 1/2)? ab=c^2+4？ ( 1/2)? Ab, so A 2+B 2 = C 2.

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Proof 3 (Zhao Shuang's String Diagram Proof):

Make four congruent triangles, a and b are right angles and c is hypotenuse, and put them together as shown in the figure below.

It is easy to get that quadrilateral ABCD and quadrilateral EFGH are both squares.

∴ area of square ABCD = area of four right triangles+area of square EFGH.

∴c^2=4？ ( 1/2)? Ab+(b-a) 2, finishing a 2+b 2 = c 2.

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Certificate 4 (Presidential Certificate):

As shown in the figure below.

It is easy to get that △CDE is an isosceles right triangle.

∴ area of trapezoidal ABCD = area of two right triangles+area of an isosceles triangle.

∴ 1/2? (a+b)？ (a+b)=2？ ( 1/2)? ab+( 1/2)？ C 2, sorted to A 2+B 2 = C 2.

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Certificate 5 (Mei Wending Certificate):

Make four congruent triangles, where A and B are right angles and C is hypotenuse, and put them together as shown in the figure below to make DEF on the same straight line. When crossing point C, CI is perpendicular to DF, and DF crosses point I. ..

Quadrilateral Abegg, Quadrilateral CBDI and Quadrilateral FGHI are all squares.

∴ area of polygon EGHCB = area of square ABEG-area of two right triangles.

And the area of polygon EGHCB = the area of square CBDI+the area of square FGHI-the areas of two right triangles.

∴ area of square ABEG = area of square CBDI+area of square FGHI

∴c？ =a？ +b？

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Proof 6 (Proof to Mindful):

Make two congruent triangles, A and B are right angles, C is hypotenuse, and make a square with a side length of C. Put them together as shown in the figure below, so that E, A and C are on the same straight line.

QP⊥AC at point Q, AC at point P.

The vertical line segments passing through F and B respectively are QP, and the intersection points are M and N respectively.

The quadrilateral ABQF is a square.

Using congruent triangles's judgment theorem, we can get the corner edge.

△ AEF△ QMF△ BNQ, then the problem is transformed into Mei Wending proof.

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Proof 7 (Euclid Proof):

In a right triangle with right angles A and B and hypotenuse C, make squares with sides A, B and C respectively, as shown in the figure below. Connect FB and CD, cross point C, do CN⊥DE at point E, and do AB at point M.

∵af=ac,ab=ad,∠fab=∠cad,∴△fab≌△cad(sas)

And delta area △FAB = delta area =△CAD = (? )? Ac sine (90 +∠CAB)=()a?

∫△CAD and right angle AMND have the same bottom height.

∴ The area of rectangular AMND is twice that of δ△CAD, that is, A?

Similarly, the area of rectangular BMNE is B?

Area of square ADEB = area of rectangular AMND+area of rectangular BMNE.

∴c？ =a？ +b？

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Proof 8 (similar triangles nature certificate)

As shown below, in the right triangle ABC, AC=b, BC=a, AB = C, ∠ ACB = 90, let point C be the CD perpendicular to AB, and point D intersect AB.

∠∠BDC =∠BCA = 90 degrees, ∠ B = ∠ B.

∴△BDC∽△BCA

∴BD∶BC=BC∶BA

∴BC？ =BD？ barium

Can I get AC in the same way? =AD？ ab blood type

∴BC？ +AC？ =BD？ BA+AD？ AB=(BD+AD)？ AB=AB？ That is, one? +b？ =c？

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Evidence 9 (Yang Zuomei's proof):

Make two congruent right-angled triangles, and let their two right-angled sides be A and B (B >; A) Make a square with a hypotenuse length of c and a side length of c, and spell it as shown in the figure below. Point A is AG⊥AC, point G is DF, point H is AG..b B is BI⊥AG, and the vertical foot is point I. Passing through point E, EJ is perpendicular to the extension line of CB, and its vertical foot is point J. EJ passes through AG at point K and DB at point L.

∠∠BAE = 90∠GAC = 90 ∴∠eak=∠bac

∵GA⊥AC,BC⊥AC

∴GA∥BC

∵EJ⊥BC

∴EJ⊥GA

∴∠eka =∠c = 90° and AE = AB = C.

∴△EAK≌△BAC(AAS)

∴EK=a,KA=b

It is easy to conclude that the quadrilateral BCAI is a rectangle.

∴AI=a,KI=b-a

∫△BAC?△EDF

∴△EAK≌△EDF

∴∠FED=∠KEA

∴∠FEK=90

∴ quadrilateral EFGK is a square, while quadrilateral DGIB is a right trapezoid.

If the number of the area is represented by a number (as shown in the figure), the area of a square with a side length of c is

c？ =S 1+S2+S3+S4+S5 ①

∵S8+S3+S4=？【b+(b-a)】？ [a+(b-a)]

=b？ -? ab，S5=S8+S9

∴S3+S4=b？ -? ab-S8=b？ -S 1-S8②

Substitute ② into ① to get.

c？ =S 1+S2+b？ -S 1-S8+S8+S9

=b？ +S2+S9

=b？ +a？

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Method 10 (Li Rui's proof):

Let the lengths of two right angles of a right triangle be a and b (b >; A) The length of the hypotenuse is C. Make three squares with side lengths of A, B and C, and put them together according to the following figure, so that the AEG is collinear at three points, the Q point is GM⊥AG, the intersection point is m, and the number of areas is represented by numbers.

∠∠TBE =∠ABH = 90

∴∠TBH=∠EBA

∵∠T=∠BEA=90，BT=BE=b

∴△HBT≌△ABE(ASA)

∴HT=AE=a,GH=GT-HT=b-a

∠GHF+∠BHT=90，∠TBH+∠BHT=90

∴∠GHF=∠TBH=∠DBC

∫BD = BE-ED = b-a，

∠G=∠BDC=90

∴△GHF≌△DBC(ASA),S7=S2

It can be seen from ∠ BAQ = ∠ BEA = 90 that ∠ABE=∠QAM.

∫AB = AQ = c

∴△ABE≌△QAM(AAS)

∴△QAM≌△HBT,S5=S8

At the same time, there are AR=AE=QM=a, and ∠QFM and ∠ACR are complementary angles of ∠GHF and ∠DBC respectively.

∴∠QFM=∠ACR

∠∠R =∠FMQ = 90°

∴△FMQ≌△CRA(AAS),S4=S6

∵c？ =S 1+S2+S3+S4+S5，

Answer? =S 1+S6，b？ =S3+S7+S8

S7=S2，S8=S5，S4=S6

∴a？ +b？ = s 1+S6+S3+S7+S8 = s 1+S4+S3+S2+S5 = c？

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Prove 1 1 (proved by tangent theorem):

In the right triangle ABC, ∠ ACB = 90, AC=b, AB=c, BC=a, draw a circle with B as the center, A as the radius, AB intersecting with point D, and AB's extension line intersecting with point E.

According to the tangent theorem (the tangent and secant of a circle are drawn from a point outside the circle, and the tangent length is the median term in the ratio of the secant to the length of two lines from the point where the secant intersects the circle), we can get: AC? =AD？ Automatic exposure device

∴b？ =(c-a)(c+a)=c？ -a？

∴a？ +b？ =c？

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Prove 12 (proved by multi-column meter theorem):

In the right triangle ABC, let BC=a, AC=b, hypotenuse AB=c, point A is AD∨CB, and point B is BD∨CA, then the quadrilateral ACBD is a rectangle inscribed with a unique circle.

According to DOM column theorem (the product of diagonal lines of quadrangles inscribed in a circle is equal to the sum of products of two opposite sides), we can get:

AB？ DC=DB？ AC+AD？ Civil band

∫AB = DC = c，DB=AC=b，AD=CB=a

∴c？ =b？ +a？

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Proof 13 (proof of inscribed circle of right triangle);

In Rt△AB=c, let right-angled side BC=a, AC=b, hypotenuse AB=c .. be the inscribed circle ⊙O of Rt△ABC, and the tangent points are D, E and F respectively, as shown in the figure below, and let the radius of circle O be R.

∫AB = AF+BF，CB=BD+CD，AC=AE+CE

∴ AC+CB-AB = (AE+CE)+(BD+CD)-(AF+BF) = CE+CD = 2r, that is, a+b-c=2r.

∴a+b=2r+c

(a+b)？ =(2r+c)？

Answer? +b？ +2ab=4(r？ +rc)+c？

∫S△ABC =？ abdominal muscle

∴4S△ABC=2ab

∫S△ABC = S△AOB+S△BOC+S△AOC =？ cr+？ ar+？ br=？ (a+b+c)r=？ (2r+c+c)r=r？ +rc

∴4(r？ +rc)=2ab

∴a？ +b？ +2ab=2ab+c？

∴a？ +b？ =c？

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Law of proof 14 (proof by reduction to absurdity):

In Rt△AB=c, let right-angled side BC=a, AC=b, hypotenuse AB = C. Point C is CD⊥AB, and vertical foot is point D, as shown in the following figure.

Suppose a? +b？ ≠c？ , which is AC? +BC？ ≠AB？

And then by AB? = ab ab = ab (ad+BD) = ab ad+ab BD。

AC？ ≠ ab ad or BC ≠AB BD

That is, AD: AC ≠ AC: AB or BD: BC ≠ BC: AB.

In △ADC and △ACB,

∠∠A =∠A

∴ If AD: AC ≠ AC: AB，∠ADC≦ACB。

In delta △CBD and delta △ACB.

∠∠B =∠B

If BD: BC ≠ BC: AB, then ∠CDB≦ACB.

∠∠ACB = 90°

∴∠adc≠90∠ CDB ≠90

This contradicts CD⊥AB, so the assumption is not valid.

∴a？ +b？ =c？

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Evidence 15 (Simpson evidence):

A right triangle has A and B as right angles and C as hypotenuse. Make a square with a side length of a+B.

Divide the square ABCD into several parts as shown in the upper left figure. The area of the square ABCD is

(a+b)？ =a？ +b？ +2ab

Divide the square ABCD into several parts, as shown on the right above, and the area of the square ABCD is

(a+b)？ =4x？ ab+c？ =2ab+c？

∴a？ +b？ +2ab=2ab+c？

∴a？ +b？ =c？

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Evidence 16 (Chen Jie evidence):

Let the lengths of two right angles of a right triangle be a and b (b >; A), the length of the hypotenuse is C. Make two squares with sides A and B, and put them in the shape as shown in the figure, so that E, H and M are in a straight line. Use numbers to represent the area code, as shown in the figure below.

If ED = a is intercepted at EH = b, and DA and DC are connected, then ad = c.

∫EM = EH+HM = b+ a，ED = a

∴DM = em――ed =(b+a)――a = b

∠∠cmd = 90，CM = a，∠ aed = 90，AE = b。

∴rtδaed≠rtδDMC(SAS)

∴ ∠EAD = ∠MDC，DC = AD = c

∠∠ADE+∠ADC+∠MDC = 180，∠ADE + ∠MDC = ∠ADE + ∠EAD = 90

∴∠ADC = 90°

∴ is abcd, CB∨da, then the quadrilateral ABCD is a square with a side length of C.

∠∠BAF+∠FAD =∠DAE+∠FAD = 90

∴ ∠BAF=∠DAE. Links FB, inδABF and ADE.

AB = AD = c，AE = AF = b，∠BAF=∠DAE