Geometry find length given altitude - Is my understanding correct?
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Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
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add a comment |
$begingroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
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You are correct, any particular reason for you to question your solution?
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– EvanHehehe
Dec 5 '18 at 3:49
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I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
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– OnePieceFan123
Dec 5 '18 at 3:53
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If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
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– EvanHehehe
Dec 5 '18 at 3:58
1
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You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
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– EvanHehehe
Dec 5 '18 at 4:01
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Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
add a comment |
$begingroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
$endgroup$
Question: For the the triangle $ABC$ , $BC=10$ and $AB=AC=13$. Given that $AD$ is a median and $BE$ is an altitude, find what $DE$ is equal to.
Well I construct the triable and I know that AD breaks BC into two equal parts of $5$. And that both $BE$ and $AD$ form right triangles.
If I look at triangle $BCE$, with angle $E$ being $90$ degrees, how can I say that $DE=DC=5=BD$
That's the answer, but I'm unsure as to why. I was leaning to thales theorem, where I can see $BDC$ as a diameter of a circle, that has angle $E$ being $90$ degrees. So they're all radius on the circle? Thus equal to 5?
geometry euclidean-geometry
geometry euclidean-geometry
edited Dec 5 '18 at 5:36
random123
1,2601720
1,2601720
asked Dec 5 '18 at 3:26
OnePieceFan123OnePieceFan123
62
62
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You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
add a comment |
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41
add a comment |
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$begingroup$
You are correct, any particular reason for you to question your solution?
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:49
$begingroup$
I was wondering if my rationale behind using thales is correct? I'm not sure exactly why, as I drew the picture to scale and DE just looks equal to BD and DC. Does Thales here only hold if angle E is equal to 90?
$endgroup$
– OnePieceFan123
Dec 5 '18 at 3:53
$begingroup$
If E was not 90 degrees, then the line BC would not be the diameter of $triangle{BCE}$'s circumcircle.
$endgroup$
– EvanHehehe
Dec 5 '18 at 3:58
1
$begingroup$
You can see proofs for what you used (the center of the circumcircle of a right triangle lies on its hypotenuse.) here
$endgroup$
– EvanHehehe
Dec 5 '18 at 4:01
$begingroup$
Ahh thank you so much! I understand now!
$endgroup$
– OnePieceFan123
Dec 5 '18 at 4:41