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

Prove that lines intersecting parallel similar triangles are concurrentDoes proving that two lines are parallel require a postulate?proving that $BC' parallel B'C$Without using angle measure how do I prove two lines are parallel to the same line are parallel to each other?Two congruent segments does have the same length?Two triangles cirumcribed a conic problemShow that two parallel lines have the same direction vector from a different definition of parallel lines.Proof: Two triangles have the same ratio of length for each corresponding side then they are similarIf the heights of two triangles are proportional then prove that they are similiarIf ratio of sides of two triangles is constant then the triangles have the same angles

5

1

$begingroup$

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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asked 9 hours ago

BanBan

653

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
  $endgroup$
  – coffeemath
  8 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
  7 hours ago
  

  
  
  
  

add a comment | 

5

1

$begingroup$

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

asked 9 hours ago

BanBan

653

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
  $endgroup$
  – coffeemath
  8 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
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

asked 9 hours ago

BanBan

653

$endgroup$

share|cite|improve this question

asked 9 hours ago

BanBan

653

share|cite|improve this question

asked 9 hours ago

BanBan

653

asked 9 hours ago

BanBan

653

asked 9 hours ago

BanBan

653

3 Answers
3

active

oldest

votes

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

answered 8 hours ago

AstaulpheAstaulphe

665

$endgroup$

add a comment | 

5

$begingroup$

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

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

add a comment | 

5

$begingroup$

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

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633

$endgroup$

add a comment | 

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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

answered 8 hours ago

AstaulpheAstaulphe

665

$endgroup$

add a comment | 

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

answered 8 hours ago

AstaulpheAstaulphe

665

$endgroup$

add a comment | 

$begingroup$

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

answered 8 hours ago

AstaulpheAstaulphe

665

$endgroup$

share|cite|improve this answer

answered 8 hours ago

AstaulpheAstaulphe

665

answered 8 hours ago

AstaulpheAstaulphe

665

answered 8 hours ago

AstaulpheAstaulphe

665

5

$begingroup$

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

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

add a comment | 

5

$begingroup$

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

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

add a comment | 

$begingroup$

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

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

answered 8 hours ago

Ethan BolkerEthan Bolker

45.7k553120

5

$begingroup$

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

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633

$endgroup$

add a comment | 

5

$begingroup$

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

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633

$endgroup$

add a comment | 

$begingroup$

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

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633

$endgroup$

share|cite|improve this answer

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633

answered 8 hours ago

eyeballfrogeyeballfrog

7,184633