Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are...












5












$begingroup$


enter image description here



Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.



I've tried proving by contradiction:



Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.



If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.



If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.



On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.



How can I show that the last case is false?






geometry euclidean-geometry







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  • $begingroup$
    But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
    $endgroup$
    – coffeemath
    11 hours ago










  • $begingroup$
    Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
    $endgroup$
    – Ban
    10 hours ago
















5












$begingroup$


enter image description here



Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.



I've tried proving by contradiction:



Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.



If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.



If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.



On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.



How can I show that the last case is false?






geometry euclidean-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
    $endgroup$
    – coffeemath
    11 hours ago










  • $begingroup$
    Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
    $endgroup$
    – Ban
    10 hours ago














5












5








5


1



$begingroup$


enter image description here



Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.



I've tried proving by contradiction:



Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.



If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.



If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.



On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.



How can I show that the last case is false?






geometry euclidean-geometry







share|cite|improve this question









$endgroup$




enter image description here



Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.



I've tried proving by contradiction:



Suppose they are not congruent but have the same perimeter, then either
|AC|$neq$|A'C| or |BC|$neq$ |B'C'|.
Let's say |AC|$neq$|A'C'|, and suppose that |AC| $lt$ |A'C'|.



If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.



If |BC| $gt$ |B'C'| then |A'C'| + |B'C'| $gt$ |AC| + |BC| which is false because their perimeters should be equal.



On the last possible case, |BC|$gt$|B'C'| I got stuck. I can't find a way to show that it is false.



How can I show that the last case is false?






geometry euclidean-geometry




geometry euclidean-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 12 hours ago









BanBan

653




653












  • $begingroup$
    But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
    $endgroup$
    – coffeemath
    11 hours ago










  • $begingroup$
    Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
    $endgroup$
    – Ban
    10 hours ago


















  • $begingroup$
    But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
    $endgroup$
    – coffeemath
    11 hours ago










  • $begingroup$
    Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
    $endgroup$
    – Ban
    10 hours ago
















$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
11 hours ago




$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
11 hours ago












$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
10 hours ago




$begingroup$
Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
$endgroup$
– Ban
10 hours ago










3 Answers
3






active

oldest

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6












$begingroup$

Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.



The easiest way to uncover your last case is using the ellipse argument.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.






    share|cite|improve this answer









    $endgroup$





















      6












      $begingroup$

      As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
      $$
      A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
      $$
      and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

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        6












        $begingroup$

        Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
        Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.



        The easiest way to uncover your last case is using the ellipse argument.






        share|cite|improve this answer









        $endgroup$


















          6












          $begingroup$

          Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
          Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.



          The easiest way to uncover your last case is using the ellipse argument.






          share|cite|improve this answer









          $endgroup$
















            6












            6








            6





            $begingroup$

            Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
            Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.



            The easiest way to uncover your last case is using the ellipse argument.






            share|cite|improve this answer









            $endgroup$



            Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse.
            Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.



            The easiest way to uncover your last case is using the ellipse argument.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 11 hours ago









            AstaulpheAstaulphe

            665




            665























                6












                $begingroup$

                Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.






                    share|cite|improve this answer









                    $endgroup$



                    Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 11 hours ago









                    Ethan BolkerEthan Bolker

                    45.7k553120




                    45.7k553120























                        6












                        $begingroup$

                        As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
                        $$
                        A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
                        $$
                        and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.






                        share|cite|improve this answer









                        $endgroup$


















                          6












                          $begingroup$

                          As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
                          $$
                          A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
                          $$
                          and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.






                          share|cite|improve this answer









                          $endgroup$
















                            6












                            6








                            6





                            $begingroup$

                            As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
                            $$
                            A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
                            $$
                            and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.






                            share|cite|improve this answer









                            $endgroup$



                            As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
                            $$
                            A = frac{1}{4}sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}
                            $$
                            and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 11 hours ago









                            eyeballfrogeyeballfrog

                            7,204633




                            7,204633






























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