Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are...
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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
$begingroup$
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 9 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 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
asked 9 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 9 hours ago
BanBan
653
asked 9 hours ago
BanBan
653
asked 9 hours ago
BanBan
653
asked 9 hours ago
BanBan
653
asked 9 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
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
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 8 hours ago
AstaulpheAstaulphe
665
$endgroup$
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 8 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 8 hours ago
eyeballfrogeyeballfrog
7,184633
$endgroup$
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 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.
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.
answered 8 hours ago
AstaulpheAstaulphe
665
$endgroup$
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 8 hours ago
AstaulpheAstaulphe
665
answered 8 hours ago
AstaulpheAstaulphe
665
answered 8 hours ago
AstaulpheAstaulphe
665
answered 8 hours ago
AstaulpheAstaulphe
665
665
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 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 $AB$.
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 $AB$.
answered 8 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 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
answered 8 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 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 $ABC cong A'B'C'$.
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 $ABC cong A'B'C'$.
answered 8 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 8 hours ago
eyeballfrogeyeballfrog
7,184633
answered 8 hours ago
eyeballfrogeyeballfrog
7,184633
answered 8 hours ago
eyeballfrogeyeballfrog
7,184633
answered 8 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
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