Gogeometry Problem 242


From Gogeomety

 

p242_equilateral_triangle

 

Solution

Using :

Solution

242

  • ΔACB’ and ΔBCA’ are equilateral => ∠ACB’ =∠BCA’ = Π/3
  • ∠ACB = ∠ACB’ – ∠B’CB = Π/3 – ∠B’CB
  • ∠B’CA’ = ∠BCA’ – ∠B’CB = Π/3 – ∠B’CB
  • =>∠ACB =∠B’CA’
  • ΔACB is congruent to ΔB’CA’ (SAS) => BA=B’A’
  • In the same way:
  • ΔCAB’ and ΔBAC’ are equilateral => ∠CAB’ =∠BAC’ = Π/3
  • ∠CAB = ∠CAB’ – ∠BAB’ = Π/3 – ∠BAB’
  • ∠B’AC’ = ∠BAC’ – ∠BAB’ = Π/3 – ∠BAB’
  • =>∠ CAB =∠ B’AC’
  • ΔCAB is congruent to ΔB’AC’ (SAS) => BC=B’C’
  • BA=B’A’ and BA=BC’ (ΔABC’ equilateral) => B’A’=BC’
  • BC=B’C’ and BC=BA’ (ΔBCA’ equilateral) => B’C’=BA’
  • => B’A’=BC’ and B’C’=BA’ => B’A’BC’ is a parallelogram

Laissez un commentaire

Votre adresse e-mail ne sera pas publiée. Les champs obligatoires sont indiqués avec *