Pythagoras theorem


From Gogeometry

p265_pythagorean_theorem

Solution

Using :

Solution

Pythagoras1

  • Define a square ABCD
  • Define E in AB, AE=a and EB=b
  • Define F in BC such as BF=b and FC=a
  • [ABCD]=[HIGD]+[EBFI]+[AEIH]+[FCGI]
  • [AEIH]=[FCGI]=ab=>[ABCD]=a^2+b^2+2ab=(a+b)^2

Pythagoras2

  • Cut [AEIH] in 2 triangles T1=AEI and T2=AHI
  • Cut [FCGI] in 2 triangles T3=FCG and T4=FIG
  • AEI, AHI, FCG and FIG are congruent with surface T=ab/2
  • Define c=AI=FG

Pythagoras3

  • [ABCD]=a^2+b^2+4T

Pythagoras4

  • Translate T2 such as HI becomes DG
  • Translate T1 such as EI becomes BF
  • Translate T4 such AS GI becomes A’A
  • By symetry, A’F’FG is a square => [A’F’FG]=c^2
  • [ABCD] =4T+[A’F’FG] =4T+c^2
  • But [ABCD]=a^2+b^2+4T
  • Therefore a^2+b^2=c^2

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