Geometry proof problem (high school) [closed] The Next CEO of Stack OverflowTriangle geometry - synthetic proofGeometric proof with a isosceles triangleProof in Geometry. Is this solution correct?Proving $triangle DCA$ is isosceles if point $D$ satisfies $angle DCA=angle DBC=30^circ$ in right isosceles $triangle ABC$High School Trigonometry Problem (I'm ashamed)Deductive geometrySolution to a-concentric-circle-problemKiselev's geometry Problem 67: In an isosceles triangle, two medians/bisectors/altitudes are congruentA simple quadrilateral is a parallelogram if:Geometry problem (proving a relation between sides of a triangle)
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Geometry proof problem (high school) [closed]
The Next CEO of Stack OverflowTriangle geometry - synthetic proofGeometric proof with a isosceles triangleProof in Geometry. Is this solution correct?Proving $triangle DCA$ is isosceles if point $D$ satisfies $angle DCA=angle DBC=30^circ$ in right isosceles $triangle ABC$High School Trigonometry Problem (I'm ashamed)Deductive geometrySolution to a-concentric-circle-problemKiselev's geometry Problem 67: In an isosceles triangle, two medians/bisectors/altitudes are congruentA simple quadrilateral is a parallelogram if:Geometry problem (proving a relation between sides of a triangle)
$begingroup$
I have an upcoming chapter test and this was one of the practice problems. Can someone guide me?
Given: Isosceles $triangle ABC$ with $AB$ congruent to $AC$; $AD$ is not a median of $triangle ABC$.
Prove: $AD$ does not bisect $angle A$.
My idea is to use proof by contradiction but I'm not sure if my thinking is right.
geometry euclidean-geometry plane-geometry
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
asked Mar 22 '13 at 23:00
PersonPerson
4731616
$endgroup$
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos Mar 21 at 15:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos
add a comment |
$begingroup$
I have an upcoming chapter test and this was one of the practice problems. Can someone guide me?
Given: Isosceles $triangle ABC$ with $AB$ congruent to $AC$; $AD$ is not a median of $triangle ABC$.
Prove: $AD$ does not bisect $angle A$.
My idea is to use proof by contradiction but I'm not sure if my thinking is right.
geometry euclidean-geometry plane-geometry
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
asked Mar 22 '13 at 23:00
PersonPerson
4731616
$endgroup$
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos Mar 21 at 15:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos
$begingroup$
Is $D$ necessarily a point on $BC$?
$endgroup$
– Sp3000
Mar 22 '13 at 23:09
$begingroup$
yes it is Sp3000
$endgroup$
– Person
Mar 22 '13 at 23:12
add a comment |
$begingroup$
I have an upcoming chapter test and this was one of the practice problems. Can someone guide me?
Given: Isosceles $triangle ABC$ with $AB$ congruent to $AC$; $AD$ is not a median of $triangle ABC$.
Prove: $AD$ does not bisect $angle A$.
My idea is to use proof by contradiction but I'm not sure if my thinking is right.
geometry euclidean-geometry plane-geometry
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
asked Mar 22 '13 at 23:00
PersonPerson
4731616
$endgroup$
I have an upcoming chapter test and this was one of the practice problems. Can someone guide me?
Given: Isosceles $triangle ABC$ with $AB$ congruent to $AC$; $AD$ is not a median of $triangle ABC$.
Prove: $AD$ does not bisect $angle A$.
My idea is to use proof by contradiction but I'm not sure if my thinking is right.
geometry euclidean-geometry plane-geometry
geometry euclidean-geometry plane-geometry
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
asked Mar 22 '13 at 23:00
PersonPerson
4731616
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
asked Mar 22 '13 at 23:00
PersonPerson
4731616
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
edited Mar 20 at 1:00
Lee David Chung Lin
4,47841242
4,47841242
asked Mar 22 '13 at 23:00
PersonPerson
4731616
asked Mar 22 '13 at 23:00
PersonPerson
4731616
asked Mar 22 '13 at 23:00
PersonPerson
4731616
4731616
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos Mar 21 at 15:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos Mar 21 at 15:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, Thomas Shelby, José Carlos Santos
$begingroup$
Is $D$ necessarily a point on $BC$?
$endgroup$
– Sp3000
Mar 22 '13 at 23:09
$begingroup$
yes it is Sp3000
$endgroup$
– Person
Mar 22 '13 at 23:12
add a comment |
$begingroup$
Is $D$ necessarily a point on $BC$?
$endgroup$
– Sp3000
Mar 22 '13 at 23:09
$begingroup$
yes it is Sp3000
$endgroup$
– Person
Mar 22 '13 at 23:12
$begingroup$
Is $D$ necessarily a point on $BC$?
$endgroup$
– Sp3000
Mar 22 '13 at 23:09
$begingroup$
Is $D$ necessarily a point on $BC$?
$endgroup$
– Sp3000
Mar 22 '13 at 23:09
$begingroup$
yes it is Sp3000
$endgroup$
– Person
Mar 22 '13 at 23:12
$begingroup$
yes it is Sp3000
$endgroup$
– Person
Mar 22 '13 at 23:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
What you can do, and this might be what you meant, is to prove the contrapositive, which is kind of like a proof by contradiction, but is more direct. We take as given the isosceles triangle with $AB cong AC$, with $D$ on $BC$, prove that
Assume $AD$ bisects $angle A,$ then prove that AD is a median of $triangle ABC$.
If you can do this, you will have proven
(I) IF $AD$ bisects $angle A,$, THEN $AD$ is a median of $triangle ABC$.
Proving (I) is equivalent to proving, with the initial "givens", that
(II) if $AD$ is NOT a median of $triangle ABC$, then AD does not bisect $angle A$.
Do you think you can do this? (Think SAS).
answered Mar 22 '13 at 23:09
NamasteNamaste
1
$endgroup$
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
|
show 1 more comment
$begingroup$
I would prove it using the contrapositive.
You want to prove this: If AD is not a median of $triangle ABC$
(which is the same as D does not bisect BC),
then AD does not bisect $angle A$.
The contrapositive, which is logically equivalent is this:
If AD bisects $angle A$
then AD is a median of $triangle ABC$
(which is the same as D bisects BC).
This should be easy to prove
(SAS).
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
If you want to proof by contradiction, then you can do this.
Assume that $AD$ is not median and bisects $angle BAC$, as it's shown on the picture below
then, according to SAS ($AB = AC$ by statement of the problem, $angle BAD = angle DAC$ by assumption and $AD$ is common edge) $triangle BAD cong triangle ADC$ which means $BD = DC$ as well, which in turn means that $D$ is median. Contradiction.
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you can do, and this might be what you meant, is to prove the contrapositive, which is kind of like a proof by contradiction, but is more direct. We take as given the isosceles triangle with $AB cong AC$, with $D$ on $BC$, prove that
Assume $AD$ bisects $angle A,$ then prove that AD is a median of $triangle ABC$.
If you can do this, you will have proven
(I) IF $AD$ bisects $angle A,$, THEN $AD$ is a median of $triangle ABC$.
Proving (I) is equivalent to proving, with the initial "givens", that
(II) if $AD$ is NOT a median of $triangle ABC$, then AD does not bisect $angle A$.
Do you think you can do this? (Think SAS).
answered Mar 22 '13 at 23:09
NamasteNamaste
1
$endgroup$
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
|
show 1 more comment
$begingroup$
What you can do, and this might be what you meant, is to prove the contrapositive, which is kind of like a proof by contradiction, but is more direct. We take as given the isosceles triangle with $AB cong AC$, with $D$ on $BC$, prove that
Assume $AD$ bisects $angle A,$ then prove that AD is a median of $triangle ABC$.
If you can do this, you will have proven
(I) IF $AD$ bisects $angle A,$, THEN $AD$ is a median of $triangle ABC$.
Proving (I) is equivalent to proving, with the initial "givens", that
(II) if $AD$ is NOT a median of $triangle ABC$, then AD does not bisect $angle A$.
Do you think you can do this? (Think SAS).
answered Mar 22 '13 at 23:09
NamasteNamaste
1
$endgroup$
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
|
show 1 more comment
$begingroup$
What you can do, and this might be what you meant, is to prove the contrapositive, which is kind of like a proof by contradiction, but is more direct. We take as given the isosceles triangle with $AB cong AC$, with $D$ on $BC$, prove that
Assume $AD$ bisects $angle A,$ then prove that AD is a median of $triangle ABC$.
If you can do this, you will have proven
(I) IF $AD$ bisects $angle A,$, THEN $AD$ is a median of $triangle ABC$.
Proving (I) is equivalent to proving, with the initial "givens", that
(II) if $AD$ is NOT a median of $triangle ABC$, then AD does not bisect $angle A$.
Do you think you can do this? (Think SAS).
answered Mar 22 '13 at 23:09
NamasteNamaste
1
$endgroup$
What you can do, and this might be what you meant, is to prove the contrapositive, which is kind of like a proof by contradiction, but is more direct. We take as given the isosceles triangle with $AB cong AC$, with $D$ on $BC$, prove that
Assume $AD$ bisects $angle A,$ then prove that AD is a median of $triangle ABC$.
If you can do this, you will have proven
(I) IF $AD$ bisects $angle A,$, THEN $AD$ is a median of $triangle ABC$.
Proving (I) is equivalent to proving, with the initial "givens", that
(II) if $AD$ is NOT a median of $triangle ABC$, then AD does not bisect $angle A$.
Do you think you can do this? (Think SAS).
answered Mar 22 '13 at 23:09
NamasteNamaste
1
answered Mar 22 '13 at 23:09
NamasteNamaste
1
answered Mar 22 '13 at 23:09
NamasteNamaste
1
answered Mar 22 '13 at 23:09
NamasteNamaste
1
1
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
|
show 1 more comment
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
Note: The contrapositive of "IF P then Q" is "If NOT Q, then NOT P" and the two statements are logically equivalent.
$endgroup$
– Namaste
Mar 22 '13 at 23:15
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
If I start out saying that $AD$ does bisect $angle A$ then I can show that $triangle ADB$ and $triangle ADC$ are congruent by SAS.But then I would need a contradiction that shows that the assumption is not compatible with the given. We are doing indirect proofs. I've never done a proof by contrapositive.
$endgroup$
– Person
Mar 22 '13 at 23:16
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Well then your assumption would be assuming BOTH that $AD$ is not a median of $triangle ABC$, and assuming that $AD$ does bisect $angle A$. Then with your conclusions, you'll know BC = DC, (that D is the midpoint of BC), and you'll find a contradiction - something that is incompatible with "$AD$ is not a median of $triangle ABC$."
$endgroup$
– Namaste
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
Thanks I understand what you're saying
$endgroup$
– Person
Mar 22 '13 at 23:20
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
$begingroup$
You're welcome. Does this make sense. I think you know exactly what you need to do. So yes, your thinking/intuition was "spot on".
$endgroup$
– Namaste
Mar 22 '13 at 23:29
|
show 1 more comment
$begingroup$
I would prove it using the contrapositive.
You want to prove this: If AD is not a median of $triangle ABC$
(which is the same as D does not bisect BC),
then AD does not bisect $angle A$.
The contrapositive, which is logically equivalent is this:
If AD bisects $angle A$
then AD is a median of $triangle ABC$
(which is the same as D bisects BC).
This should be easy to prove
(SAS).
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
I would prove it using the contrapositive.
You want to prove this: If AD is not a median of $triangle ABC$
(which is the same as D does not bisect BC),
then AD does not bisect $angle A$.
The contrapositive, which is logically equivalent is this:
If AD bisects $angle A$
then AD is a median of $triangle ABC$
(which is the same as D bisects BC).
This should be easy to prove
(SAS).
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
$endgroup$
add a comment |
$begingroup$
I would prove it using the contrapositive.
You want to prove this: If AD is not a median of $triangle ABC$
(which is the same as D does not bisect BC),
then AD does not bisect $angle A$.
The contrapositive, which is logically equivalent is this:
If AD bisects $angle A$
then AD is a median of $triangle ABC$
(which is the same as D bisects BC).
This should be easy to prove
(SAS).
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
$endgroup$
I would prove it using the contrapositive.
You want to prove this: If AD is not a median of $triangle ABC$
(which is the same as D does not bisect BC),
then AD does not bisect $angle A$.
The contrapositive, which is logically equivalent is this:
If AD bisects $angle A$
then AD is a median of $triangle ABC$
(which is the same as D bisects BC).
This should be easy to prove
(SAS).
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
answered Mar 22 '13 at 23:12
marty cohenmarty cohen
74.9k549130
74.9k549130
add a comment |
add a comment |
$begingroup$
If you want to proof by contradiction, then you can do this.
Assume that $AD$ is not median and bisects $angle BAC$, as it's shown on the picture below
then, according to SAS ($AB = AC$ by statement of the problem, $angle BAD = angle DAC$ by assumption and $AD$ is common edge) $triangle BAD cong triangle ADC$ which means $BD = DC$ as well, which in turn means that $D$ is median. Contradiction.
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
$endgroup$
add a comment |
$begingroup$
If you want to proof by contradiction, then you can do this.
Assume that $AD$ is not median and bisects $angle BAC$, as it's shown on the picture below
then, according to SAS ($AB = AC$ by statement of the problem, $angle BAD = angle DAC$ by assumption and $AD$ is common edge) $triangle BAD cong triangle ADC$ which means $BD = DC$ as well, which in turn means that $D$ is median. Contradiction.
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
$endgroup$
add a comment |
$begingroup$
If you want to proof by contradiction, then you can do this.
Assume that $AD$ is not median and bisects $angle BAC$, as it's shown on the picture below
then, according to SAS ($AB = AC$ by statement of the problem, $angle BAD = angle DAC$ by assumption and $AD$ is common edge) $triangle BAD cong triangle ADC$ which means $BD = DC$ as well, which in turn means that $D$ is median. Contradiction.
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
$endgroup$
If you want to proof by contradiction, then you can do this.
Assume that $AD$ is not median and bisects $angle BAC$, as it's shown on the picture below
then, according to SAS ($AB = AC$ by statement of the problem, $angle BAD = angle DAC$ by assumption and $AD$ is common edge) $triangle BAD cong triangle ADC$ which means $BD = DC$ as well, which in turn means that $D$ is median. Contradiction.
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
answered Mar 22 '13 at 23:26
KasterKaster
9,08221730
9,08221730
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$begingroup$
Is $D$ necessarily a point on $BC$?
$endgroup$
– Sp3000
Mar 22 '13 at 23:09
$begingroup$
yes it is Sp3000
$endgroup$
– Person
Mar 22 '13 at 23:12