Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are...
$begingroup$
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 7 hours ago
BanBan
653
$endgroup$
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 7 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
6 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
5 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
asked 7 hours ago
BanBan
653
$endgroup$
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 7 hours ago
BanBan
653
asked 7 hours ago
BanBan
653
asked 7 hours ago
BanBan
653
asked 7 hours ago
BanBan
653
asked 7 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
6 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
5 hours 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
6 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
5 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
6 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
6 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
5 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
5 hours 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 6 hours ago
AstaulpheAstaulphe
565
$endgroup$
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 6 hours 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 6 hours ago
eyeballfrogeyeballfrog
7,184633
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add a comment |
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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 6 hours ago
AstaulpheAstaulphe
565
$endgroup$
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 6 hours ago
AstaulpheAstaulphe
565
$endgroup$
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 6 hours ago
AstaulpheAstaulphe
565
$endgroup$
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 6 hours ago
AstaulpheAstaulphe
565
answered 6 hours ago
AstaulpheAstaulphe
565
answered 6 hours ago
AstaulpheAstaulphe
565
answered 6 hours ago
AstaulpheAstaulphe
565
565
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 6 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 $AB$.
answered 6 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 $AB$.
answered 6 hours 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 6 hours ago
Ethan BolkerEthan Bolker
45.7k553120
answered 6 hours ago
Ethan BolkerEthan Bolker
45.7k553120
answered 6 hours ago
Ethan BolkerEthan Bolker
45.7k553120
answered 6 hours 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 6 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 $ABC cong A'B'C'$.
answered 6 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 $ABC cong A'B'C'$.
answered 6 hours ago
eyeballfrogeyeballfrog
7,184633
$endgroup$
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 6 hours ago
eyeballfrogeyeballfrog
7,184633
answered 6 hours ago
eyeballfrogeyeballfrog
7,184633
answered 6 hours ago
eyeballfrogeyeballfrog
7,184633
answered 6 hours ago
eyeballfrogeyeballfrog
7,184633
7,184633
add a comment |
add a comment |
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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
6 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
5 hours ago