Midpoints of CeviansProve midpoints collinearWhy are $triangle ABC$ and $triangle CDE$ similar?Prove triangles formed by two midpoints and an altitude are congruentFind an angle in a triangle with ceviansProve that quadrilateral $ADOE$ is cyclicMid-sections and anglesProve that lines passing through the midpoints of sides of a triangle and the midpoints of cevians are also concurrentA geometry problem asking to prove point of intersection of $2$ cevians lies on circumcircle of the triangle.(South Africa 2014) Finding angles in an obtuse triangleProblem involving triangle with characterization of midpoints
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Midpoints of Cevians
Prove midpoints collinearWhy are $triangle ABC$ and $triangle CDE$ similar?Prove triangles formed by two midpoints and an altitude are congruentFind an angle in a triangle with ceviansProve that quadrilateral $ADOE$ is cyclicMid-sections and anglesProve that lines passing through the midpoints of sides of a triangle and the midpoints of cevians are also concurrentA geometry problem asking to prove point of intersection of $2$ cevians lies on circumcircle of the triangle.(South Africa 2014) Finding angles in an obtuse triangleProblem involving triangle with characterization of midpoints
$begingroup$
In triangle $triangle ABC$, $D$ and $E$ lie on sides $CA$ and $AB$ such that $BE = 6$ and $CD = 10$. Let $M$ and $N$ be the midpoints of segments $BD$ and $CE$, respectively. If $MN = 7$, then what is the measure of $angle BAC$?
I'm not sure how to start this, I can't find any similar or parallel lines in the figure. I can construct the trapezoids but I do not know how to utilize them.
geometry triangles
edited Mar 24 at 17:05
Dr. Mathva
3,190630
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
$endgroup$
add a comment |
$begingroup$
In triangle $triangle ABC$, $D$ and $E$ lie on sides $CA$ and $AB$ such that $BE = 6$ and $CD = 10$. Let $M$ and $N$ be the midpoints of segments $BD$ and $CE$, respectively. If $MN = 7$, then what is the measure of $angle BAC$?
I'm not sure how to start this, I can't find any similar or parallel lines in the figure. I can construct the trapezoids but I do not know how to utilize them.
geometry triangles
edited Mar 24 at 17:05
Dr. Mathva
3,190630
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
$endgroup$
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
Mar 24 at 16:31
$begingroup$
sorry no picture was given in the problem
$endgroup$
– SuperMage1
Mar 24 at 16:33
$begingroup$
You could make one with geogebra for instance; that would make the problem more attractive... And please consider using $LaTeX$
$endgroup$
– Dr. Mathva
Mar 24 at 16:40
add a comment |
$begingroup$
In triangle $triangle ABC$, $D$ and $E$ lie on sides $CA$ and $AB$ such that $BE = 6$ and $CD = 10$. Let $M$ and $N$ be the midpoints of segments $BD$ and $CE$, respectively. If $MN = 7$, then what is the measure of $angle BAC$?
I'm not sure how to start this, I can't find any similar or parallel lines in the figure. I can construct the trapezoids but I do not know how to utilize them.
geometry triangles
edited Mar 24 at 17:05
Dr. Mathva
3,190630
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
$endgroup$
In triangle $triangle ABC$, $D$ and $E$ lie on sides $CA$ and $AB$ such that $BE = 6$ and $CD = 10$. Let $M$ and $N$ be the midpoints of segments $BD$ and $CE$, respectively. If $MN = 7$, then what is the measure of $angle BAC$?
I'm not sure how to start this, I can't find any similar or parallel lines in the figure. I can construct the trapezoids but I do not know how to utilize them.
geometry triangles
geometry triangles
edited Mar 24 at 17:05
Dr. Mathva
3,190630
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
edited Mar 24 at 17:05
Dr. Mathva
3,190630
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
edited Mar 24 at 17:05
Dr. Mathva
3,190630
edited Mar 24 at 17:05
Dr. Mathva
3,190630
edited Mar 24 at 17:05
Dr. Mathva
3,190630
3,190630
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
asked Mar 24 at 16:18
SuperMage1SuperMage1
950211
950211
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
Mar 24 at 16:31
$begingroup$
sorry no picture was given in the problem
$endgroup$
– SuperMage1
Mar 24 at 16:33
$begingroup$
You could make one with geogebra for instance; that would make the problem more attractive... And please consider using $LaTeX$
$endgroup$
– Dr. Mathva
Mar 24 at 16:40
add a comment |
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
Mar 24 at 16:31
$begingroup$
sorry no picture was given in the problem
$endgroup$
– SuperMage1
Mar 24 at 16:33
$begingroup$
You could make one with geogebra for instance; that would make the problem more attractive... And please consider using $LaTeX$
$endgroup$
– Dr. Mathva
Mar 24 at 16:40
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
Mar 24 at 16:31
$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
Mar 24 at 16:31
$begingroup$
sorry no picture was given in the problem
$endgroup$
– SuperMage1
Mar 24 at 16:33
$begingroup$
sorry no picture was given in the problem
$endgroup$
– SuperMage1
Mar 24 at 16:33
$begingroup$
You could make one with geogebra for instance; that would make the problem more attractive... And please consider using $LaTeX$
$endgroup$
– Dr. Mathva
Mar 24 at 16:40
$begingroup$
You could make one with geogebra for instance; that would make the problem more attractive... And please consider using $LaTeX$
$endgroup$
– Dr. Mathva
Mar 24 at 16:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $A=E=D$, then triangle $MAN$ has sides $3$, $5$ and $7$. From the cosine rule it follows that
$$
cosangle MAN=3^2+5^5-7^2over 2cdot 3cdot 5=-1over2,
quadtextthat is:quad angle MAN=angle BAC=120°.
$$
This is a particular case, but the problem implies that the same solution must hold also in the general case.
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $A=E=D$, then triangle $MAN$ has sides $3$, $5$ and $7$. From the cosine rule it follows that
$$
cosangle MAN=3^2+5^5-7^2over 2cdot 3cdot 5=-1over2,
quadtextthat is:quad angle MAN=angle BAC=120°.
$$
This is a particular case, but the problem implies that the same solution must hold also in the general case.
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
add a comment |
$begingroup$
If $A=E=D$, then triangle $MAN$ has sides $3$, $5$ and $7$. From the cosine rule it follows that
$$
cosangle MAN=3^2+5^5-7^2over 2cdot 3cdot 5=-1over2,
quadtextthat is:quad angle MAN=angle BAC=120°.
$$
This is a particular case, but the problem implies that the same solution must hold also in the general case.
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
add a comment |
$begingroup$
If $A=E=D$, then triangle $MAN$ has sides $3$, $5$ and $7$. From the cosine rule it follows that
$$
cosangle MAN=3^2+5^5-7^2over 2cdot 3cdot 5=-1over2,
quadtextthat is:quad angle MAN=angle BAC=120°.
$$
This is a particular case, but the problem implies that the same solution must hold also in the general case.
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
$endgroup$
If $A=E=D$, then triangle $MAN$ has sides $3$, $5$ and $7$. From the cosine rule it follows that
$$
cosangle MAN=3^2+5^5-7^2over 2cdot 3cdot 5=-1over2,
quadtextthat is:quad angle MAN=angle BAC=120°.
$$
This is a particular case, but the problem implies that the same solution must hold also in the general case.
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
edited Mar 29 at 8:13
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
answered Mar 24 at 17:42
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
add a comment |
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
what do you mean by A =E = D?
$endgroup$
– SuperMage1
Mar 29 at 4:01
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
$begingroup$
See my edit: I had written by mistake $E$ and $D$ instead of $M$ and $N$, I corrected and added a picture.
$endgroup$
– Aretino
Mar 29 at 8:18
add a comment |
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$begingroup$
Could you provide a picture?
$endgroup$
– Dr. Mathva
Mar 24 at 16:31
$begingroup$
sorry no picture was given in the problem
$endgroup$
– SuperMage1
Mar 24 at 16:33
$begingroup$
You could make one with geogebra for instance; that would make the problem more attractive... And please consider using $LaTeX$
$endgroup$
– Dr. Mathva
Mar 24 at 16:40