Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are...
Which models of the Boeing 737 are still in production?
How much RAM could one put in a typical 80386 setup?
How to know the difference between two ciphertexts without key stream in stream ciphers
What does it mean to describe someone as a butt steak?
Why don't electron-positron collisions release infinite energy?
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?
To string or not to string
US citizen flying to France today and my passport expires in less than 2 months
Smoothness of finite-dimensional functional calculus
How does one intimidate enemies without having the capacity for violence?
Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?
How is it possible to have an ability score that is less than 3?
Is it possible to do 50 km distance without any previous training?
Unknown notation: What do three bars mean?
Why is 150k or 200k jobs considered good when there's 300k+ births a month?
How to format long polynomial?
Why does Kotter return in Welcome Back Kotter?
Equivalence principle before Einstein
What's the point of deactivating Num Lock on login screens?
Test whether all array elements are factors of a number
Important Resources for Dark Age Civilizations?
How does strength of boric acid solution increase in presence of salicylic acid?
Adding span tags within wp_list_pages list items
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.
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
$begingroup$
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
asked 1 hour ago
BanBan
553
$endgroup$
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
asked 1 hour ago
BanBan
553
$endgroup$
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
18 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
asked 1 hour ago
BanBan
553
$endgroup$
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
asked 1 hour ago
BanBan
553
asked 1 hour ago
BanBan
553
asked 1 hour ago
BanBan
553
asked 1 hour ago
BanBan
553
asked 1 hour ago
BanBan
553
553
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
18 mins ago
add a comment |
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
18 mins ago
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
18 mins ago
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
18 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered 18 mins ago
AstaulpheAstaulphe
265
$endgroup$
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 $AB$.
answered 11 mins ago
Ethan BolkerEthan Bolker
45.7k553120
$endgroup$
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 $ABC cong A'B'C'$.
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
$endgroup$
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
});
}
});
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
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
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 18 mins ago
AstaulpheAstaulphe
265
$endgroup$
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.
answered 18 mins ago
AstaulpheAstaulphe
265
$endgroup$
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.
answered 18 mins ago
AstaulpheAstaulphe
265
$endgroup$
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.
answered 18 mins ago
AstaulpheAstaulphe
265
answered 18 mins ago
AstaulpheAstaulphe
265
answered 18 mins ago
AstaulpheAstaulphe
265
answered 18 mins ago
AstaulpheAstaulphe
265
265
add a comment |
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 $AB$.
answered 11 mins 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 $AB$.
answered 11 mins 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 $AB$.
answered 11 mins ago
Ethan BolkerEthan Bolker
45.7k553120
$endgroup$
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 $AB$.
answered 11 mins ago
Ethan BolkerEthan Bolker
45.7k553120
answered 11 mins ago
Ethan BolkerEthan Bolker
45.7k553120
answered 11 mins ago
Ethan BolkerEthan Bolker
45.7k553120
answered 11 mins ago
Ethan BolkerEthan Bolker
45.7k553120
45.7k553120
add a comment |
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 $ABC cong A'B'C'$.
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
$endgroup$
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 $ABC cong A'B'C'$.
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
$endgroup$
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 $ABC cong A'B'C'$.
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
$endgroup$
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 $ABC cong A'B'C'$.
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
answered 9 mins ago
eyeballfrogeyeballfrog
7,134633
7,134633
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent.
$endgroup$
– coffeemath
18 mins ago