Mixing
What is the name of this result about isosceles triangles?
Clash Royale CLAN TAG#URR8PPP
up vote
26
down vote
favorite
$Delta ABC$ is isosceles with $AB=AC=p$. $D$ is a point on $BC$ where $AD=q$, $BD=u$ and $CD=v$.
Then the following holds.
$$p^2=q^2+uv$$
I would like to know whether this result has a name.
geometry euclidean-geometry triangle
edited Sep 8 at 18:17
Jam
4,36111330
asked Sep 5 at 9:34
Yuta
61429
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
26
down vote
favorite
$Delta ABC$ is isosceles with $AB=AC=p$. $D$ is a point on $BC$ where $AD=q$, $BD=u$ and $CD=v$.
Then the following holds.
$$p^2=q^2+uv$$
I would like to know whether this result has a name.
geometry euclidean-geometry triangle
edited Sep 8 at 18:17
Jam
4,36111330
asked Sep 5 at 9:34
Yuta
61429
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
If you have time ( still in high school or freshman) read "Geometry revisted" by coxeter. It will really help .
– blue boy
Sep 5 at 10:14
1
Thanks for the suggestion. I will check it out.
– Yuta
Sep 5 at 10:24
1
One could also name it as "trivial" ;)
– Jan
Sep 5 at 13:08
2
@Jan Indeed, just saying "trivial" is a great way to change the subject. It's like "left as an exercise for the reader".
– Sneftel
Sep 6 at 10:24
@Jan But why leave it as folklore?
– BCLC
Sep 6 at 14:18
add a comment |Â
up vote
26
down vote
favorite
up vote
26
down vote
favorite
$Delta ABC$ is isosceles with $AB=AC=p$. $D$ is a point on $BC$ where $AD=q$, $BD=u$ and $CD=v$.
Then the following holds.
$$p^2=q^2+uv$$
I would like to know whether this result has a name.
geometry euclidean-geometry triangle
edited Sep 8 at 18:17
Jam
4,36111330
asked Sep 5 at 9:34
Yuta
61429
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$Delta ABC$ is isosceles with $AB=AC=p$. $D$ is a point on $BC$ where $AD=q$, $BD=u$ and $CD=v$.
Then the following holds.
$$p^2=q^2+uv$$
I would like to know whether this result has a name.
geometry euclidean-geometry triangle
edited Sep 8 at 18:17
Jam
4,36111330
asked Sep 5 at 9:34
Yuta
61429
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 8 at 18:17
Jam
4,36111330
edited Sep 8 at 18:17
Jam
4,36111330
edited Sep 8 at 18:17
Jam
4,36111330
4,36111330
asked Sep 5 at 9:34
Yuta
61429
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 5 at 9:34
Yuta
61429
asked Sep 5 at 9:34
Yuta
61429
61429
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yuta is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
If you have time ( still in high school or freshman) read "Geometry revisted" by coxeter. It will really help .
– blue boy
Sep 5 at 10:14
1
Thanks for the suggestion. I will check it out.
– Yuta
Sep 5 at 10:24
1
One could also name it as "trivial" ;)
– Jan
Sep 5 at 13:08
2
@Jan Indeed, just saying "trivial" is a great way to change the subject. It's like "left as an exercise for the reader".
– Sneftel
Sep 6 at 10:24
@Jan But why leave it as folklore?
– BCLC
Sep 6 at 14:18
add a comment |Â
1
If you have time ( still in high school or freshman) read "Geometry revisted" by coxeter. It will really help .
– blue boy
Sep 5 at 10:14
1
Thanks for the suggestion. I will check it out.
– Yuta
Sep 5 at 10:24
1
One could also name it as "trivial" ;)
– Jan
Sep 5 at 13:08
2
@Jan Indeed, just saying "trivial" is a great way to change the subject. It's like "left as an exercise for the reader".
– Sneftel
Sep 6 at 10:24
@Jan But why leave it as folklore?
– BCLC
Sep 6 at 14:18
1
1
If you have time ( still in high school or freshman) read "Geometry revisted" by coxeter. It will really help .
– blue boy
Sep 5 at 10:14
If you have time ( still in high school or freshman) read "Geometry revisted" by coxeter. It will really help .
– blue boy
Sep 5 at 10:14
1
1
Thanks for the suggestion. I will check it out.
– Yuta
Sep 5 at 10:24
Thanks for the suggestion. I will check it out.
– Yuta
Sep 5 at 10:24
1
1
One could also name it as "trivial" ;)
– Jan
Sep 5 at 13:08
One could also name it as "trivial" ;)
– Jan
Sep 5 at 13:08
2
2
@Jan Indeed, just saying "trivial" is a great way to change the subject. It's like "left as an exercise for the reader".
– Sneftel
Sep 6 at 10:24
@Jan Indeed, just saying "trivial" is a great way to change the subject. It's like "left as an exercise for the reader".
– Sneftel
Sep 6 at 10:24
@Jan But why leave it as folklore?
– BCLC
Sep 6 at 14:18
@Jan But why leave it as folklore?
– BCLC
Sep 6 at 14:18
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
39
down vote
accepted
Stewart's theorem
Let $a$, $b$, and $c$ are the length of the sides of the triangles. Let $d$ be the length of the cevian to the side of length $a$.  If the cevian divides the side of length $a$ into two segments of length $m$ and $n$ with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that
$b^2m +c^2n=a(d^2+mn)$
Note
Apollonius's theorem is a special case of this theorem.
edited Sep 5 at 10:46
Glorfindel
3,21371729
answered Sep 5 at 10:08
blue boy
1,095513
2
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
add a comment |Â
up vote
7
down vote
Edit: I forgot some factors $frac12$.
Derivation:
$$h^2 = p^2 - frac14(u+v)^2 ,$$
$$q^2 = h^2 + frac14(u-v)^2 ,$$
where $h$ is the height perpendicular to $BC$. Therefore,
$$p^2 = h^2 + frac14(u+v)^2 = q^2 - frac14(u-v)^2 + frac14(u+v)^2 = q^2 + uv .$$
I don't know if it has a name, but it is certainly a nice formula!
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
add a comment |Â
up vote
6
down vote
Besides Stewart's theorem, Ptolemy's theorem also gives this one as an immediate consequence:
$
deflen#1overline#1
defangangle
defparaparallel
deftritriangle
$
Let $E$ be the reflection of $A$ about the perpendicular bisector of $CD$. Then clearly $lenCE = q$ and $lenDE = p$. Also $DE para AB$, so $tri ABD equiv tri DEA$, and hence $lenAE = u$. Finally $ADCE$ is cyclic, so Ptolemy's theorem gives the desired result.
answered Sep 5 at 12:33
user21820
36.2k440140
add a comment |Â
up vote
5
down vote
This result (with several proofs) appears in a note by L. Hoehn in The Mathematical Gazette (March 2000 pp 71-73). The author suggests, plausibly, that it must have been found many times, but notes that "it doesn't seem to appear in the mathematical literature". He does not refer to or propose any name for the result.
answered Sep 5 at 14:57
Adam Bailey
1,9241218
add a comment |Â
up vote
2
down vote
It can be seen as a very straight-forward corollary of the intersecting chords theorem. If $c$ is a circle with center $A$ and passing through $B$ and $C$, $c$ has radius $p$. Then $BC$ is a chord of $c$ and line $AD$ intersect $c$ forming a diameter, say at points $E$ and $F$ (suppose $E$ is the point closer to $D$). Then $ED=p-q$ and $FD=p+q$. Your result follows from the intersecting chord theorem: $EDcdot FD=BDcdot CD$.
answered Sep 5 at 19:18
olaphus
563
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olaphus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
Let $M$ be the midpoint of $BC$. Let $MD = d$.
It is easy to see that $v - u = 2d$ and so $v = u + 2d$.
By Pythagoras
$$p^2- q^2
= MC^2 - MD^2
= (MC + MD)(MC - MD)
= [(u + v)/2 + d][(u + v)/2 - d]
= [(2u + 2d)/2 + d] [(2u + 2d)/2 - d]
= (u + 2d)u = vu.$$
Hence $p^2 = q^2 + uv$.
edited Sep 5 at 22:27
Alan Muniz
1,411622
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
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Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
39
down vote
accepted
Stewart's theorem
Let $a$, $b$, and $c$ are the length of the sides of the triangles. Let $d$ be the length of the cevian to the side of length $a$.  If the cevian divides the side of length $a$ into two segments of length $m$ and $n$ with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that
$b^2m +c^2n=a(d^2+mn)$
Note
Apollonius's theorem is a special case of this theorem.
edited Sep 5 at 10:46
Glorfindel
3,21371729
answered Sep 5 at 10:08
blue boy
1,095513
2
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
add a comment |Â
up vote
39
down vote
accepted
Stewart's theorem
Let $a$, $b$, and $c$ are the length of the sides of the triangles. Let $d$ be the length of the cevian to the side of length $a$.  If the cevian divides the side of length $a$ into two segments of length $m$ and $n$ with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that
$b^2m +c^2n=a(d^2+mn)$
Note
Apollonius's theorem is a special case of this theorem.
edited Sep 5 at 10:46
Glorfindel
3,21371729
answered Sep 5 at 10:08
blue boy
1,095513
2
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
add a comment |Â
up vote
39
down vote
accepted
up vote
39
down vote
accepted
Stewart's theorem
Let $a$, $b$, and $c$ are the length of the sides of the triangles. Let $d$ be the length of the cevian to the side of length $a$.  If the cevian divides the side of length $a$ into two segments of length $m$ and $n$ with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that
$b^2m +c^2n=a(d^2+mn)$
Note
Apollonius's theorem is a special case of this theorem.
edited Sep 5 at 10:46
Glorfindel
3,21371729
answered Sep 5 at 10:08
blue boy
1,095513
Stewart's theorem
Let $a$, $b$, and $c$ are the length of the sides of the triangles. Let $d$ be the length of the cevian to the side of length $a$.  If the cevian divides the side of length $a$ into two segments of length $m$ and $n$ with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that
$b^2m +c^2n=a(d^2+mn)$
Note
Apollonius's theorem is a special case of this theorem.
edited Sep 5 at 10:46
Glorfindel
3,21371729
answered Sep 5 at 10:08
blue boy
1,095513
edited Sep 5 at 10:46
Glorfindel
3,21371729
edited Sep 5 at 10:46
Glorfindel
3,21371729
edited Sep 5 at 10:46
Glorfindel
3,21371729
3,21371729
answered Sep 5 at 10:08
blue boy
1,095513
answered Sep 5 at 10:08
blue boy
1,095513
answered Sep 5 at 10:08
blue boy
1,095513
1,095513
2
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
add a comment |Â
2
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
2
2
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
Actually I am just wondering if there is a more general theorem that includes the result that I posted and Apollonius's theorem. And here comes exactly what I am looking for. Thank you so much!
– Yuta
Sep 5 at 10:22
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
This has the useful mnemonic $man+dad=bmb+cnc$
– Display name
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Displayname Are there full (or related) names for bmb and cnc?
– Mick
2 days ago
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
@Mick A man and his dad put a bomb in the sink.
– Display name
yesterday
add a comment |Â
up vote
7
down vote
Edit: I forgot some factors $frac12$.
Derivation:
$$h^2 = p^2 - frac14(u+v)^2 ,$$
$$q^2 = h^2 + frac14(u-v)^2 ,$$
where $h$ is the height perpendicular to $BC$. Therefore,
$$p^2 = h^2 + frac14(u+v)^2 = q^2 - frac14(u-v)^2 + frac14(u+v)^2 = q^2 + uv .$$
I don't know if it has a name, but it is certainly a nice formula!
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
add a comment |Â
up vote
7
down vote
Edit: I forgot some factors $frac12$.
Derivation:
$$h^2 = p^2 - frac14(u+v)^2 ,$$
$$q^2 = h^2 + frac14(u-v)^2 ,$$
where $h$ is the height perpendicular to $BC$. Therefore,
$$p^2 = h^2 + frac14(u+v)^2 = q^2 - frac14(u-v)^2 + frac14(u+v)^2 = q^2 + uv .$$
I don't know if it has a name, but it is certainly a nice formula!
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Edit: I forgot some factors $frac12$.
Derivation:
$$h^2 = p^2 - frac14(u+v)^2 ,$$
$$q^2 = h^2 + frac14(u-v)^2 ,$$
where $h$ is the height perpendicular to $BC$. Therefore,
$$p^2 = h^2 + frac14(u+v)^2 = q^2 - frac14(u-v)^2 + frac14(u+v)^2 = q^2 + uv .$$
I don't know if it has a name, but it is certainly a nice formula!
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
Edit: I forgot some factors $frac12$.
Derivation:
$$h^2 = p^2 - frac14(u+v)^2 ,$$
$$q^2 = h^2 + frac14(u-v)^2 ,$$
where $h$ is the height perpendicular to $BC$. Therefore,
$$p^2 = h^2 + frac14(u+v)^2 = q^2 - frac14(u-v)^2 + frac14(u+v)^2 = q^2 + uv .$$
I don't know if it has a name, but it is certainly a nice formula!
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
edited Sep 5 at 10:04
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
answered Sep 5 at 9:39
Daniel Robert-Nicoud
19.8k33495
19.8k33495
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
add a comment |Â
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
I obtained the result in exactly the same way!
– Yuta
Sep 5 at 10:06
add a comment |Â
up vote
6
down vote
Besides Stewart's theorem, Ptolemy's theorem also gives this one as an immediate consequence:
$
deflen#1overline#1
defangangle
defparaparallel
deftritriangle
$
Let $E$ be the reflection of $A$ about the perpendicular bisector of $CD$. Then clearly $lenCE = q$ and $lenDE = p$. Also $DE para AB$, so $tri ABD equiv tri DEA$, and hence $lenAE = u$. Finally $ADCE$ is cyclic, so Ptolemy's theorem gives the desired result.
answered Sep 5 at 12:33
user21820
36.2k440140
add a comment |Â
up vote
6
down vote
Besides Stewart's theorem, Ptolemy's theorem also gives this one as an immediate consequence:
$
deflen#1overline#1
defangangle
defparaparallel
deftritriangle
$
Let $E$ be the reflection of $A$ about the perpendicular bisector of $CD$. Then clearly $lenCE = q$ and $lenDE = p$. Also $DE para AB$, so $tri ABD equiv tri DEA$, and hence $lenAE = u$. Finally $ADCE$ is cyclic, so Ptolemy's theorem gives the desired result.
answered Sep 5 at 12:33
user21820
36.2k440140
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Besides Stewart's theorem, Ptolemy's theorem also gives this one as an immediate consequence:
$
deflen#1overline#1
defangangle
defparaparallel
deftritriangle
$
Let $E$ be the reflection of $A$ about the perpendicular bisector of $CD$. Then clearly $lenCE = q$ and $lenDE = p$. Also $DE para AB$, so $tri ABD equiv tri DEA$, and hence $lenAE = u$. Finally $ADCE$ is cyclic, so Ptolemy's theorem gives the desired result.
answered Sep 5 at 12:33
user21820
36.2k440140
Besides Stewart's theorem, Ptolemy's theorem also gives this one as an immediate consequence:
$
deflen#1overline#1
defangangle
defparaparallel
deftritriangle
$
Let $E$ be the reflection of $A$ about the perpendicular bisector of $CD$. Then clearly $lenCE = q$ and $lenDE = p$. Also $DE para AB$, so $tri ABD equiv tri DEA$, and hence $lenAE = u$. Finally $ADCE$ is cyclic, so Ptolemy's theorem gives the desired result.
answered Sep 5 at 12:33
user21820
36.2k440140
answered Sep 5 at 12:33
user21820
36.2k440140
answered Sep 5 at 12:33
user21820
36.2k440140
answered Sep 5 at 12:33
user21820
36.2k440140
36.2k440140
add a comment |Â
add a comment |Â
up vote
5
down vote
This result (with several proofs) appears in a note by L. Hoehn in The Mathematical Gazette (March 2000 pp 71-73). The author suggests, plausibly, that it must have been found many times, but notes that "it doesn't seem to appear in the mathematical literature". He does not refer to or propose any name for the result.
answered Sep 5 at 14:57
Adam Bailey
1,9241218
add a comment |Â
up vote
5
down vote
This result (with several proofs) appears in a note by L. Hoehn in The Mathematical Gazette (March 2000 pp 71-73). The author suggests, plausibly, that it must have been found many times, but notes that "it doesn't seem to appear in the mathematical literature". He does not refer to or propose any name for the result.
answered Sep 5 at 14:57
Adam Bailey
1,9241218
add a comment |Â
up vote
5
down vote
up vote
5
down vote
This result (with several proofs) appears in a note by L. Hoehn in The Mathematical Gazette (March 2000 pp 71-73). The author suggests, plausibly, that it must have been found many times, but notes that "it doesn't seem to appear in the mathematical literature". He does not refer to or propose any name for the result.
answered Sep 5 at 14:57
Adam Bailey
1,9241218
This result (with several proofs) appears in a note by L. Hoehn in The Mathematical Gazette (March 2000 pp 71-73). The author suggests, plausibly, that it must have been found many times, but notes that "it doesn't seem to appear in the mathematical literature". He does not refer to or propose any name for the result.
answered Sep 5 at 14:57
Adam Bailey
1,9241218
answered Sep 5 at 14:57
Adam Bailey
1,9241218
answered Sep 5 at 14:57
Adam Bailey
1,9241218
answered Sep 5 at 14:57
Adam Bailey
1,9241218
1,9241218
add a comment |Â
add a comment |Â
up vote
2
down vote
It can be seen as a very straight-forward corollary of the intersecting chords theorem. If $c$ is a circle with center $A$ and passing through $B$ and $C$, $c$ has radius $p$. Then $BC$ is a chord of $c$ and line $AD$ intersect $c$ forming a diameter, say at points $E$ and $F$ (suppose $E$ is the point closer to $D$). Then $ED=p-q$ and $FD=p+q$. Your result follows from the intersecting chord theorem: $EDcdot FD=BDcdot CD$.
answered Sep 5 at 19:18
olaphus
563
New contributor
olaphus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
It can be seen as a very straight-forward corollary of the intersecting chords theorem. If $c$ is a circle with center $A$ and passing through $B$ and $C$, $c$ has radius $p$. Then $BC$ is a chord of $c$ and line $AD$ intersect $c$ forming a diameter, say at points $E$ and $F$ (suppose $E$ is the point closer to $D$). Then $ED=p-q$ and $FD=p+q$. Your result follows from the intersecting chord theorem: $EDcdot FD=BDcdot CD$.
answered Sep 5 at 19:18
olaphus
563
New contributor
olaphus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It can be seen as a very straight-forward corollary of the intersecting chords theorem. If $c$ is a circle with center $A$ and passing through $B$ and $C$, $c$ has radius $p$. Then $BC$ is a chord of $c$ and line $AD$ intersect $c$ forming a diameter, say at points $E$ and $F$ (suppose $E$ is the point closer to $D$). Then $ED=p-q$ and $FD=p+q$. Your result follows from the intersecting chord theorem: $EDcdot FD=BDcdot CD$.
answered Sep 5 at 19:18
olaphus
563
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It can be seen as a very straight-forward corollary of the intersecting chords theorem. If $c$ is a circle with center $A$ and passing through $B$ and $C$, $c$ has radius $p$. Then $BC$ is a chord of $c$ and line $AD$ intersect $c$ forming a diameter, say at points $E$ and $F$ (suppose $E$ is the point closer to $D$). Then $ED=p-q$ and $FD=p+q$. Your result follows from the intersecting chord theorem: $EDcdot FD=BDcdot CD$.
answered Sep 5 at 19:18
olaphus
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edited Sep 6 at 21:46
answered Sep 5 at 19:18
olaphus
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answered Sep 5 at 19:18
olaphus
563
answered Sep 5 at 19:18
olaphus
563
563
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up vote
0
down vote
Let $M$ be the midpoint of $BC$. Let $MD = d$.
It is easy to see that $v - u = 2d$ and so $v = u + 2d$.
By Pythagoras
$$p^2- q^2
= MC^2 - MD^2
= (MC + MD)(MC - MD)
= [(u + v)/2 + d][(u + v)/2 - d]
= [(2u + 2d)/2 + d] [(2u + 2d)/2 - d]
= (u + 2d)u = vu.$$
Hence $p^2 = q^2 + uv$.
edited Sep 5 at 22:27
Alan Muniz
1,411622
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
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I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
add a comment |Â
up vote
0
down vote
Let $M$ be the midpoint of $BC$. Let $MD = d$.
It is easy to see that $v - u = 2d$ and so $v = u + 2d$.
By Pythagoras
$$p^2- q^2
= MC^2 - MD^2
= (MC + MD)(MC - MD)
= [(u + v)/2 + d][(u + v)/2 - d]
= [(2u + 2d)/2 + d] [(2u + 2d)/2 - d]
= (u + 2d)u = vu.$$
Hence $p^2 = q^2 + uv$.
edited Sep 5 at 22:27
Alan Muniz
1,411622
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
New contributor
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $M$ be the midpoint of $BC$. Let $MD = d$.
It is easy to see that $v - u = 2d$ and so $v = u + 2d$.
By Pythagoras
$$p^2- q^2
= MC^2 - MD^2
= (MC + MD)(MC - MD)
= [(u + v)/2 + d][(u + v)/2 - d]
= [(2u + 2d)/2 + d] [(2u + 2d)/2 - d]
= (u + 2d)u = vu.$$
Hence $p^2 = q^2 + uv$.
edited Sep 5 at 22:27
Alan Muniz
1,411622
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
New contributor
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $M$ be the midpoint of $BC$. Let $MD = d$.
It is easy to see that $v - u = 2d$ and so $v = u + 2d$.
By Pythagoras
$$p^2- q^2
= MC^2 - MD^2
= (MC + MD)(MC - MD)
= [(u + v)/2 + d][(u + v)/2 - d]
= [(2u + 2d)/2 + d] [(2u + 2d)/2 - d]
= (u + 2d)u = vu.$$
Hence $p^2 = q^2 + uv$.
edited Sep 5 at 22:27
Alan Muniz
1,411622
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
New contributor
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 5 at 22:27
Alan Muniz
1,411622
edited Sep 5 at 22:27
Alan Muniz
1,411622
edited Sep 5 at 22:27
Alan Muniz
1,411622
1,411622
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
New contributor
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
answered Sep 5 at 21:50
Vijaya Prasad Nalluri
1
1
New contributor
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Vijaya Prasad Nalluri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
add a comment |Â
I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
I think the relation is $u - v = 2d$
– David
Sep 6 at 1:42
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
But the question asks for the name of the theorem. You don't mention that.
– Arnaud D.
Sep 6 at 11:54
add a comment |Â
Yuta is a new contributor. Be nice, and check out our Code of Conduct.
Yuta is a new contributor. Be nice, and check out our Code of Conduct.
Yuta is a new contributor. Be nice, and check out our Code of Conduct.
Yuta is a new contributor. Be nice, and check out our Code of Conduct.
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1
If you have time ( still in high school or freshman) read "Geometry revisted" by coxeter. It will really help .
– blue boy
Sep 5 at 10:14
1
Thanks for the suggestion. I will check it out.
– Yuta
Sep 5 at 10:24
1
One could also name it as "trivial" ;)
– Jan
Sep 5 at 13:08
2
@Jan Indeed, just saying "trivial" is a great way to change the subject. It's like "left as an exercise for the reader".
– Sneftel
Sep 6 at 10:24
@Jan But why leave it as folklore?
– BCLC
Sep 6 at 14:18