Axial symmetry of triangle ABC AB=A’B’, BC=B’C’, AC=A’C’ ∠ABC= ∠A’B’C’, ∠BAC=∠B’A’C’, ∠ACB= ∠A’C’B’ See also « Isometric transformation«
Harvey Littleman
Displacement of Triangle ABC AB=A’B’, BC=B’C’, AC=A’C’ ∠ABC= ∠A’B’C’, ∠BAC=∠B’A’C’, ∠ACB= ∠A’C’B’ See also « Isometric transformation«
Isometric transformation – Displacement
Rotation of triangle ABC from point O with angle α AB=A’B’, BC=B’C’, AC=A’C’ ∠ABC= ∠A’B’C’, ∠BAC=∠B’A’C’, ∠ACB= ∠A’C’B’ ∠(AB,A’B’)=∠(AC,A’C’)=∠(BC,B’C’)=α See also « Isometric transformation«
Isometric transformation – Rotation
To solve a problem, it could be necessary to add one or more points.
Additional points
Isometric transformation and combinations of Isometric transformation preserve angles and distances: displacement rotation axial symmetry central symmetry
Isometric transformation
From Gogeomety Using : Equilateral triangle Congruent triangles Parallelogram 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 […]
Gogeometry Problem 242
From Gogeometry : Using : Congruent Triangles Equilateral triangle ∠CBC’=∠ABC+∠ABC’= ∠ABC+60° ∠ABA’= ∠ABC+∠CBA’= ∠ABC+60° =>∠CBC’=∠ABA’ ΔABA’ is congruent to ΔCBC’ (SAS) Therefore AA’=CC’
Gogeometry Problem 241
From Gogeometry : Using : Congruent triangles ΔABE is congruent to ΔBCF by rotation of pi/2 => <AGF=90° Define H as the middle of AB and K as the middle of AG By construction HD//BF AH=HB=>AK=KG <AGF=90° => <AKD=90° ΔDKG is congruent to ΔDKA (SAS) =>DG=DA Therefore a=b
Gogeometry Problem 238
From Gogeometry : Using : Equilateral triangle Congruent triangles Parallelogram ΔCAB’ and ΔC’AB are equilateral => ∠CAB’=∠C’AB=Π/3 ∠BAB’=a+∠CAB’=∠BAC’+a’ => a=a’ AC’=AB and AB’=AC and a=a’=> ΔBAC and ΔB’AC’ are congruent (SAS) =>B’C’=BC But BC=BA’=>B’C’=BA’ In the same way : ∠BCA’=∠B’CA=Π/3 ∠BCA=c= Π/3 – ∠ACA’ ∠B’CA’=c’= Π/3 – ∠ACA’ =>c=c’ […]
Gogeometry Problem 240
Equilateral triangle <=> AB=AC=BC ∠CAB=∠ABC=∠BCA=Π/3=60°