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EuclideanGeometry.Example.IMO.IMO2007

Geometry Problems of 48th International Mathematical Olympiad (2007) #

Problem 2 #

Consider five points $A$, $B$, $C$, $D$, $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $l$ be a line passing through $A$, and let $l$ intersect segment $DC$ and line $BC$ at points $F$ and $G$, respectively. Suppose that $EF$ = $EG$ = $EC$. Prove that $l$ is the bisector of angle $DAB$.

Instances For

    $C$ ≠ $B$ since $ABCD$ is a nondegenerate parallelogram.

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    $D$ ≠ $C$ since $ABCD$ is a nondegenerate parallelogram.

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    $D$ ≠ $A$ since $ABCD$ is a nondegenerate parallelogram.

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    $B$ ≠ $A$ since $ABCD$ is a nondegenerate parallelogram.

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    $F$ ≠ $C$ since $F$ lies in the interior the segment $DC$.

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    Rename line $BC$ as lbc.

    Equations
    • e.lbc = LIN e.B e.C (_ : e.C e.B)
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      Rename line $DC$ as ldc.

      Equations
      • e.ldc = LIN e.D e.C (_ : e.C e.D)
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        @[inline, reducible]

        Rename the nondegenerate parallelogram $ABCD$ as prg.

        Equations
        Instances For

          $B$, $C$ and $D$ are not collinear since $ABCD$ is a nondegenerate parallelogram.

          $C$, $D$ and $A$ are not collinear since $ABCD$ is a nondegenerate parallelogram.

          $C$, $D$ and $A$ are not collinear since $ABCD$ is a nondegenerate parallelogram.

          $F$ lies on line $DC$ since $F$ lies in the interior the segment $DC$.

          Line $DC$ is not equal to line $BC$ since $B$, $C$ and $D$ are not collinear.

          Line $DC$ is not equal to line $AB$ since $A$, $B$ and $D$ are not collinear.

          Line $DC$ is not equal to line $BC$ since $B$, $C$ and $D$ are not collinear.

          Line $DC$ is parallel to line $AB$ since $ABCD$ is a nondegenerate parallelogram.

          Line $DA$ is parallel to line $BC$ since $ABCD$ is a nondegenerate parallelogram.

          $G$ ≠ $A$.

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          $G$ ≠ $F$.

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          $G$ ≠ $B$.

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          $G$ ≠ $C$.

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          $F$ lies in the interior the segment $AG$ due to the conditions of parallel lines.

          $C$ lies in the interior the segment $BG$ due to the conditions of parallel lines.

          $E$ ≠ $B$.

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          $E$ ≠ $C$.

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          $E$ ≠ $D$.

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          theorem EuclidGeom.IMO_2007_2.Setting.hbcde {P : Type u_1} [EuclidGeom.EuclideanPlane P] (e : EuclidGeom.IMO_2007_2.Setting P) :
          e.C e.B e.E = e.C e.D e.E

          ∠$CBE$ and ∠$CDE$ are congruent modulo π since $BCED$ is a cyclic quadrilateral.

          @[inline, reducible]

          Let $K$ be the midpoint of segment $CG$.

          Equations
          • e.K = { source := e.C, target := e.G }.midpoint
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            @[inline, reducible]

            Let $L$ be the midpoint of segment $FC$.

            Equations
            • e.L = { source := e.F, target := e.C }.midpoint
            Instances For

              $K$ ≠ $B$ since it is the midpoint of the nondegenerate segment $CG$.

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              $K$ ≠ $C$ since it is the midpoint of the nondegenerate segment $CG$.

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              $L$ ≠ $F$ since it is the midpoint of the nondegenerate segment $FC$.

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              $L$ ≠ $C$ since it is the midpoint of the nondegenerate segment $FC$.

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              $L$ ≠ $D$ since $L$ lies in the interior the segment $DC$.

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              $K$ is the perpendicular foot from $E$ to line $BC$.

              $L$ is the perpendicular foot from $E$ to line $DC$.

              $K$ lies on line $BC$ since it is the perpendicular foot from $E$ to line $BC$.

              $L$ lies on line $DC$ since it is the perpendicular foot from $E$ to line $DC$.

              Rename nondegenerate ▵$KBE$ as KBE.

              Equations
              • e.KBE = { val := { point₁ := e.K, point₂ := e.B, point₃ := e.E }, property := (_ : ¬EuclidGeom.Collinear e.K e.B { point₁ := e.K, point₂ := e.B, point₃ := e.E }.point₃) }
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                Rename nondegenerate ▵$LDE$ as LDE.

                Equations
                • e.LDE = { val := { point₁ := e.L, point₂ := e.D, point₃ := e.E }, property := (_ : ¬EuclidGeom.Collinear e.L e.D { point₁ := e.L, point₂ := e.D, point₃ := e.E }.point₃) }
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                  Rename nondegenerate ▵$KCE$ as KCE.

                  Equations
                  • e.KCE = { val := { point₁ := e.K, point₂ := e.C, point₃ := e.E }, property := (_ : ¬EuclidGeom.Collinear e.K e.C { point₁ := e.K, point₂ := e.C, point₃ := e.E }.point₃) }
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                    Rename nondegenerate ▵$LCE$ as LCE.

                    Equations
                    • e.LCE = { val := { point₁ := e.L, point₂ := e.C, point₃ := e.E }, property := (_ : ¬EuclidGeom.Collinear e.L e.C { point₁ := e.L, point₂ := e.C, point₃ := e.E }.point₃) }
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                      theorem EuclidGeom.IMO_2007_2.ratio_eq {P : Type u_1} [EuclidGeom.EuclideanPlane P] (e : EuclidGeom.IMO_2007_2.Setting P) :
                      dist e.E e.K / dist e.E e.L = dist e.K e.C / dist e.L e.C

                      $|EK| / |EL| = |KC| / |LC|$.

                      theorem EuclidGeom.IMO_2007_2.tri_CGF_isIsoceles {P : Type u_1} [EuclidGeom.EuclideanPlane P] (e : EuclidGeom.IMO_2007_2.Setting P) :
                      EuclidGeom.Triangle.IsIsoceles { point₁ := e.C, point₂ := e.G, point₃ := e.F }

                      ▵$CGF$ is a isosceles triangle.

                      $l$ is the bisector of angle $DAB$.