Hjdcti

Why is no PWM signal generated using Timer 2?

Why is 150k or 200k jobs considered good when there's 300k+ births a month?

Dragon forelimb placement

How to add double frame in tcolorbox?

What are these boxed doors outside store fronts in New York?

How is it possible to have an ability score that is less than 3?

What defenses are there against being summoned by the Gate spell?

Why can't I see bouncing of a switch on an oscilloscope?

"to be prejudice towards/against someone" vs "to be prejudiced against/towards someone"

Can divisibility rules for digits be generalized to sum of digits

tikz: show 0 at the axis origin

What typically incentivizes a professor to change jobs to a lower ranking university?

can i play a electric guitar through a bass amp?

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

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

Do I have a twin with permutated remainders?

Why, historically, did Gödel think CH was false?

In Japanese, what’s the difference between “Tonari ni” (となりに) and “Tsugi” (つぎ)? When would you use one over the other?

Is it legal for company to use my work email to pretend I still work there?

Watching something be written to a file live with tail

Did Shadowfax go to Valinor?

Theorems that impeded progress

Pattern match does not work in bash script

Which models of the Boeing 737 are still in production?

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

4

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 2 hours ago

BanBan

603

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

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 2 hours ago

BanBan

603

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false.
  $endgroup$
  – Ban
  36 mins 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 2 hours ago

BanBan

603

$endgroup$

share|cite|improve this question

asked 2 hours ago

BanBan

603

share|cite|improve this question

asked 2 hours ago

BanBan

603

asked 2 hours ago

BanBan

603

asked 2 hours ago

BanBan

603

3 Answers
3

active

oldest

votes

4

$begingroup$

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 1 hour ago

AstaulpheAstaulphe

465

$endgroup$

add a comment | 

3

$begingroup$

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 1 hour ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

add a comment | 

3

$begingroup$

As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac14sqrt(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)
$$
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 1 hour ago

eyeballfrogeyeballfrog

7,164633

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177306%2fshow-that-if-two-triangles-built-on-parallel-lines-with-equal-bases-have-the-sa%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

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 1 hour ago

AstaulpheAstaulphe

465

$endgroup$

add a comment | 

4

$begingroup$

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 1 hour ago

AstaulpheAstaulphe

465

$endgroup$

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 1 hour ago

AstaulpheAstaulphe

465

$endgroup$

share|cite|improve this answer

answered 1 hour ago

AstaulpheAstaulphe

465

answered 1 hour ago

AstaulpheAstaulphe

465

answered 1 hour ago

AstaulpheAstaulphe

465

3

$begingroup$

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 1 hour ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

add a comment | 

3

$begingroup$

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 1 hour 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 1 hour ago

Ethan BolkerEthan Bolker

45.7k553120

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Ethan BolkerEthan Bolker

45.7k553120

answered 1 hour ago

Ethan BolkerEthan Bolker

45.7k553120

answered 1 hour ago

Ethan BolkerEthan Bolker

45.7k553120

3

$begingroup$

As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac14sqrt(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)
$$
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 1 hour ago

eyeballfrogeyeballfrog

7,164633

$endgroup$

add a comment | 

3

$begingroup$

As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac14sqrt(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)
$$
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 1 hour ago

eyeballfrogeyeballfrog

7,164633

$endgroup$

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
$$
A = frac14sqrt(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)
$$
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 1 hour ago

eyeballfrogeyeballfrog

7,164633

$endgroup$

share|cite|improve this answer

answered 1 hour ago

eyeballfrogeyeballfrog

7,164633

answered 1 hour ago

eyeballfrogeyeballfrog

7,164633

answered 1 hour ago

eyeballfrogeyeballfrog

7,164633