Mixing
SAT Math Problem - Corresponding Angles in Similar Triangles
$begingroup$
In the following problem, why must ∠BAE ≅ ∠CED? Can't ∠BAE ≅ ∠BDE as well if you simply flip the triangle on top around?
For instance: enter image description here
geometry
asked Jan 11 at 19:15
user633879user633879
32
$endgroup$
add a comment |
$begingroup$
In the following problem, why must ∠BAE ≅ ∠CED? Can't ∠BAE ≅ ∠BDE as well if you simply flip the triangle on top around?
For instance: enter image description here
geometry
asked Jan 11 at 19:15
user633879user633879
32
$endgroup$
$begingroup$
Try to draw the line $AE$ more « horizontally » and look what happens.
$endgroup$
– Mindlack
Jan 11 at 19:31
add a comment |
$begingroup$
In the following problem, why must ∠BAE ≅ ∠CED? Can't ∠BAE ≅ ∠BDE as well if you simply flip the triangle on top around?
For instance: enter image description here
geometry
asked Jan 11 at 19:15
user633879user633879
32
$endgroup$
In the following problem, why must ∠BAE ≅ ∠CED? Can't ∠BAE ≅ ∠BDE as well if you simply flip the triangle on top around?
For instance: enter image description here
geometry
geometry
asked Jan 11 at 19:15
user633879user633879
32
asked Jan 11 at 19:15
user633879user633879
32
edited Jan 11 at 19:54
user633879
asked Jan 11 at 19:15
user633879user633879
32
asked Jan 11 at 19:15
user633879user633879
32
asked Jan 11 at 19:15
user633879user633879
32
32
$begingroup$
Try to draw the line $AE$ more « horizontally » and look what happens.
$endgroup$
– Mindlack
Jan 11 at 19:31
add a comment |
$begingroup$
Try to draw the line $AE$ more « horizontally » and look what happens.
$endgroup$
– Mindlack
Jan 11 at 19:31
$begingroup$
Try to draw the line $AE$ more « horizontally » and look what happens.
$endgroup$
– Mindlack
Jan 11 at 19:31
$begingroup$
Try to draw the line $AE$ more « horizontally » and look what happens.
$endgroup$
– Mindlack
Jan 11 at 19:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I assume it's not gived that $AB || DE$: it's not mentioned anywhere, and in the picture they aren't parallel.
In that case, it's only a matter of notation. When we write, $triangle ABD sim triangle DEF$, the order of the vertices matters, that is, $angle A = angle D$, $angle B = angle E$, and $angle C = angle F$. Similarly, in your case $angle A = angle E$, since they're written in corresponding places.
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
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add a comment |
$begingroup$
It is $angle BAE=angle AED$ since you are flipping the triangle along left-right(i.e. along a vertical axis such that it is proportional in size to the other, i.e. the angles are same and the side lengths are in the same ratio). If you're confused, its good to remember that in general
$triangle ABCsim triangle DEFRightarrowfrac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}$ also, $angle CAB=angle FDE,angle ABC=angle DEF,angle BCA=angle EFD$
answered Jan 11 at 19:38
MustangMustang
3367
$endgroup$
add a comment |
$begingroup$
If we assume that $ACE$ is a straight line (not explicitly stated, but a reasonable inference), then $∠BAE$ is the same as $∠BAC$. We're given that $Delta ABC ≅ Delta EDC$. When two triangles are given as congruent, that means that corresponding elements are congruent. It might help to write the letters in different colors, but apparently SE doesn't support that, so I'll use different typefaces: $Delta$ A B $mathbb C$ $≅ Delta$ E D $mathbb C$.
From this, we can conclude $∠$B A $mathbb C$ $≅∠$ D E $mathbb C$. Note that the typefaces match up: on the left, it's italics, bold, blackboard. On the right, it's italics, bold, blackboard.
Is $∠BAE ≅ ∠BDE$? Well, it could be. But the question doesn't ask for what could be, it asks for what must be. Writing that with my typefaces, (and keeping in mind that $AE$ and $AC$ describe the same line, as do $BD$ and $CD$) it's $∠$B A $mathbb C ≅ ∠mathbb C$ D E. The typefaces don't match up: the left is italics, bold, blackboard, but the right is blackboard, italics, bold. So while that doesn't prove that the two angles are different, it does show that their congruence isn't required.
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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$begingroup$
I assume it's not gived that $AB || DE$: it's not mentioned anywhere, and in the picture they aren't parallel.
In that case, it's only a matter of notation. When we write, $triangle ABD sim triangle DEF$, the order of the vertices matters, that is, $angle A = angle D$, $angle B = angle E$, and $angle C = angle F$. Similarly, in your case $angle A = angle E$, since they're written in corresponding places.
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
$endgroup$
add a comment |
$begingroup$
I assume it's not gived that $AB || DE$: it's not mentioned anywhere, and in the picture they aren't parallel.
In that case, it's only a matter of notation. When we write, $triangle ABD sim triangle DEF$, the order of the vertices matters, that is, $angle A = angle D$, $angle B = angle E$, and $angle C = angle F$. Similarly, in your case $angle A = angle E$, since they're written in corresponding places.
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
$endgroup$
add a comment |
$begingroup$
I assume it's not gived that $AB || DE$: it's not mentioned anywhere, and in the picture they aren't parallel.
In that case, it's only a matter of notation. When we write, $triangle ABD sim triangle DEF$, the order of the vertices matters, that is, $angle A = angle D$, $angle B = angle E$, and $angle C = angle F$. Similarly, in your case $angle A = angle E$, since they're written in corresponding places.
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
$endgroup$
I assume it's not gived that $AB || DE$: it's not mentioned anywhere, and in the picture they aren't parallel.
In that case, it's only a matter of notation. When we write, $triangle ABD sim triangle DEF$, the order of the vertices matters, that is, $angle A = angle D$, $angle B = angle E$, and $angle C = angle F$. Similarly, in your case $angle A = angle E$, since they're written in corresponding places.
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
answered Jan 11 at 19:39
Todor MarkovTodor Markov
2,201411
2,201411
add a comment |
add a comment |
$begingroup$
It is $angle BAE=angle AED$ since you are flipping the triangle along left-right(i.e. along a vertical axis such that it is proportional in size to the other, i.e. the angles are same and the side lengths are in the same ratio). If you're confused, its good to remember that in general
$triangle ABCsim triangle DEFRightarrowfrac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}$ also, $angle CAB=angle FDE,angle ABC=angle DEF,angle BCA=angle EFD$
answered Jan 11 at 19:38
MustangMustang
3367
$endgroup$
add a comment |
$begingroup$
It is $angle BAE=angle AED$ since you are flipping the triangle along left-right(i.e. along a vertical axis such that it is proportional in size to the other, i.e. the angles are same and the side lengths are in the same ratio). If you're confused, its good to remember that in general
$triangle ABCsim triangle DEFRightarrowfrac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}$ also, $angle CAB=angle FDE,angle ABC=angle DEF,angle BCA=angle EFD$
answered Jan 11 at 19:38
MustangMustang
3367
$endgroup$
add a comment |
$begingroup$
It is $angle BAE=angle AED$ since you are flipping the triangle along left-right(i.e. along a vertical axis such that it is proportional in size to the other, i.e. the angles are same and the side lengths are in the same ratio). If you're confused, its good to remember that in general
$triangle ABCsim triangle DEFRightarrowfrac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}$ also, $angle CAB=angle FDE,angle ABC=angle DEF,angle BCA=angle EFD$
answered Jan 11 at 19:38
MustangMustang
3367
$endgroup$
It is $angle BAE=angle AED$ since you are flipping the triangle along left-right(i.e. along a vertical axis such that it is proportional in size to the other, i.e. the angles are same and the side lengths are in the same ratio). If you're confused, its good to remember that in general
$triangle ABCsim triangle DEFRightarrowfrac{AB}{DE}=frac{BC}{EF}=frac{AC}{DF}$ also, $angle CAB=angle FDE,angle ABC=angle DEF,angle BCA=angle EFD$
answered Jan 11 at 19:38
MustangMustang
3367
answered Jan 11 at 19:38
MustangMustang
3367
answered Jan 11 at 19:38
MustangMustang
3367
answered Jan 11 at 19:38
MustangMustang
3367
3367
add a comment |
add a comment |
$begingroup$
If we assume that $ACE$ is a straight line (not explicitly stated, but a reasonable inference), then $∠BAE$ is the same as $∠BAC$. We're given that $Delta ABC ≅ Delta EDC$. When two triangles are given as congruent, that means that corresponding elements are congruent. It might help to write the letters in different colors, but apparently SE doesn't support that, so I'll use different typefaces: $Delta$ A B $mathbb C$ $≅ Delta$ E D $mathbb C$.
From this, we can conclude $∠$B A $mathbb C$ $≅∠$ D E $mathbb C$. Note that the typefaces match up: on the left, it's italics, bold, blackboard. On the right, it's italics, bold, blackboard.
Is $∠BAE ≅ ∠BDE$? Well, it could be. But the question doesn't ask for what could be, it asks for what must be. Writing that with my typefaces, (and keeping in mind that $AE$ and $AC$ describe the same line, as do $BD$ and $CD$) it's $∠$B A $mathbb C ≅ ∠mathbb C$ D E. The typefaces don't match up: the left is italics, bold, blackboard, but the right is blackboard, italics, bold. So while that doesn't prove that the two angles are different, it does show that their congruence isn't required.
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
$endgroup$
add a comment |
$begingroup$
If we assume that $ACE$ is a straight line (not explicitly stated, but a reasonable inference), then $∠BAE$ is the same as $∠BAC$. We're given that $Delta ABC ≅ Delta EDC$. When two triangles are given as congruent, that means that corresponding elements are congruent. It might help to write the letters in different colors, but apparently SE doesn't support that, so I'll use different typefaces: $Delta$ A B $mathbb C$ $≅ Delta$ E D $mathbb C$.
From this, we can conclude $∠$B A $mathbb C$ $≅∠$ D E $mathbb C$. Note that the typefaces match up: on the left, it's italics, bold, blackboard. On the right, it's italics, bold, blackboard.
Is $∠BAE ≅ ∠BDE$? Well, it could be. But the question doesn't ask for what could be, it asks for what must be. Writing that with my typefaces, (and keeping in mind that $AE$ and $AC$ describe the same line, as do $BD$ and $CD$) it's $∠$B A $mathbb C ≅ ∠mathbb C$ D E. The typefaces don't match up: the left is italics, bold, blackboard, but the right is blackboard, italics, bold. So while that doesn't prove that the two angles are different, it does show that their congruence isn't required.
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
$endgroup$
add a comment |
$begingroup$
If we assume that $ACE$ is a straight line (not explicitly stated, but a reasonable inference), then $∠BAE$ is the same as $∠BAC$. We're given that $Delta ABC ≅ Delta EDC$. When two triangles are given as congruent, that means that corresponding elements are congruent. It might help to write the letters in different colors, but apparently SE doesn't support that, so I'll use different typefaces: $Delta$ A B $mathbb C$ $≅ Delta$ E D $mathbb C$.
From this, we can conclude $∠$B A $mathbb C$ $≅∠$ D E $mathbb C$. Note that the typefaces match up: on the left, it's italics, bold, blackboard. On the right, it's italics, bold, blackboard.
Is $∠BAE ≅ ∠BDE$? Well, it could be. But the question doesn't ask for what could be, it asks for what must be. Writing that with my typefaces, (and keeping in mind that $AE$ and $AC$ describe the same line, as do $BD$ and $CD$) it's $∠$B A $mathbb C ≅ ∠mathbb C$ D E. The typefaces don't match up: the left is italics, bold, blackboard, but the right is blackboard, italics, bold. So while that doesn't prove that the two angles are different, it does show that their congruence isn't required.
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
$endgroup$
If we assume that $ACE$ is a straight line (not explicitly stated, but a reasonable inference), then $∠BAE$ is the same as $∠BAC$. We're given that $Delta ABC ≅ Delta EDC$. When two triangles are given as congruent, that means that corresponding elements are congruent. It might help to write the letters in different colors, but apparently SE doesn't support that, so I'll use different typefaces: $Delta$ A B $mathbb C$ $≅ Delta$ E D $mathbb C$.
From this, we can conclude $∠$B A $mathbb C$ $≅∠$ D E $mathbb C$. Note that the typefaces match up: on the left, it's italics, bold, blackboard. On the right, it's italics, bold, blackboard.
Is $∠BAE ≅ ∠BDE$? Well, it could be. But the question doesn't ask for what could be, it asks for what must be. Writing that with my typefaces, (and keeping in mind that $AE$ and $AC$ describe the same line, as do $BD$ and $CD$) it's $∠$B A $mathbb C ≅ ∠mathbb C$ D E. The typefaces don't match up: the left is italics, bold, blackboard, but the right is blackboard, italics, bold. So while that doesn't prove that the two angles are different, it does show that their congruence isn't required.
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
answered Jan 11 at 19:58
AcccumulationAcccumulation
6,9042618
6,9042618
add a comment |
add a comment |
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$begingroup$
Try to draw the line $AE$ more « horizontally » and look what happens.
$endgroup$
– Mindlack
Jan 11 at 19:31