Geometry with triangle ABC
$begingroup$
My question is...
I want to know your another solution.
or I want to know if my solution is appropriate.
and I’d appreciate some feedback on my work.
$E$ is midpoint of $overline{BC}$.
$overline{AD} : overline{DE}$ = 4:3
Find the $overline{AF} : overline{FC}$.
geometry proof-verification vectors euclidean-geometry triangle
asked Jan 24 at 15:46
mina_worldmina_world
1779
$endgroup$
|
show 1 more comment
$begingroup$
My question is...
I want to know your another solution.
or I want to know if my solution is appropriate.
and I’d appreciate some feedback on my work.
$E$ is midpoint of $overline{BC}$.
$overline{AD} : overline{DE}$ = 4:3
Find the $overline{AF} : overline{FC}$.
geometry proof-verification vectors euclidean-geometry triangle
asked Jan 24 at 15:46
mina_worldmina_world
1779
$endgroup$
5
$begingroup$
In the future, please use MathJax to format your mathematics, because this makes your question more search-engine friendly. Having said that, though, the way you've typeset your question is beautiful and very easy to read.
$endgroup$
– John
Jan 24 at 15:58
$begingroup$
@John In fact, that's why I always write the problem with MathJax in the first place. (for search).
$endgroup$
– mina_world
Jan 24 at 16:05
1
$begingroup$
@mina_world: All the text should be actual text. Someone might remember that there was some recent question involving, say, the terms "midpoint" and "collinear"; since "collinear" is trapped in the image, it doesn't get read. (This isn't the best example, but I hope you take my meaning: You want as much of your work as possible to be searchable.) As I mentioned in a comment to another item of yours: screen readers (for the visually impaired) need text. (Granted, MathJax is also problematic for screen readers, but there are occasional words in there. :)
$endgroup$
– Blue
Jan 24 at 16:22
1
$begingroup$
While I'm handing out friendly advice :) ... It would be really helpful to provide more-informative titles to your questions. "Geometry with triangle ABC", "Geometry with circle and triangle", "Geometry with triangle" don't tell the reader what to expect. Moreover, if these questions had come up in a search, the searcher wouldn't be able to tell which questions are actually relevant. (Years from now, when you decide to look back through your own questions for something that could help someone, you won't be able to tell which is which. Try to give questions titles that help "Future You".)
$endgroup$
– Blue
Jan 24 at 16:25
3
$begingroup$
@Blue I will not be satisfied with everything, but I will always do my best. Thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:31
|
show 1 more comment
$begingroup$
My question is...
I want to know your another solution.
or I want to know if my solution is appropriate.
and I’d appreciate some feedback on my work.
$E$ is midpoint of $overline{BC}$.
$overline{AD} : overline{DE}$ = 4:3
Find the $overline{AF} : overline{FC}$.
geometry proof-verification vectors euclidean-geometry triangle
asked Jan 24 at 15:46
mina_worldmina_world
1779
$endgroup$
My question is...
I want to know your another solution.
or I want to know if my solution is appropriate.
and I’d appreciate some feedback on my work.
$E$ is midpoint of $overline{BC}$.
$overline{AD} : overline{DE}$ = 4:3
Find the $overline{AF} : overline{FC}$.
geometry proof-verification vectors euclidean-geometry triangle
geometry proof-verification vectors euclidean-geometry triangle
asked Jan 24 at 15:46
mina_worldmina_world
1779
asked Jan 24 at 15:46
mina_worldmina_world
1779
edited Jan 25 at 7:21
mina_world
asked Jan 24 at 15:46
mina_worldmina_world
1779
asked Jan 24 at 15:46
mina_worldmina_world
1779
asked Jan 24 at 15:46
mina_worldmina_world
1779
1779
5
$begingroup$
In the future, please use MathJax to format your mathematics, because this makes your question more search-engine friendly. Having said that, though, the way you've typeset your question is beautiful and very easy to read.
$endgroup$
– John
Jan 24 at 15:58
$begingroup$
@John In fact, that's why I always write the problem with MathJax in the first place. (for search).
$endgroup$
– mina_world
Jan 24 at 16:05
1
$begingroup$
@mina_world: All the text should be actual text. Someone might remember that there was some recent question involving, say, the terms "midpoint" and "collinear"; since "collinear" is trapped in the image, it doesn't get read. (This isn't the best example, but I hope you take my meaning: You want as much of your work as possible to be searchable.) As I mentioned in a comment to another item of yours: screen readers (for the visually impaired) need text. (Granted, MathJax is also problematic for screen readers, but there are occasional words in there. :)
$endgroup$
– Blue
Jan 24 at 16:22
1
$begingroup$
While I'm handing out friendly advice :) ... It would be really helpful to provide more-informative titles to your questions. "Geometry with triangle ABC", "Geometry with circle and triangle", "Geometry with triangle" don't tell the reader what to expect. Moreover, if these questions had come up in a search, the searcher wouldn't be able to tell which questions are actually relevant. (Years from now, when you decide to look back through your own questions for something that could help someone, you won't be able to tell which is which. Try to give questions titles that help "Future You".)
$endgroup$
– Blue
Jan 24 at 16:25
3
$begingroup$
@Blue I will not be satisfied with everything, but I will always do my best. Thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:31
|
show 1 more comment
5
$begingroup$
In the future, please use MathJax to format your mathematics, because this makes your question more search-engine friendly. Having said that, though, the way you've typeset your question is beautiful and very easy to read.
$endgroup$
– John
Jan 24 at 15:58
$begingroup$
@John In fact, that's why I always write the problem with MathJax in the first place. (for search).
$endgroup$
– mina_world
Jan 24 at 16:05
1
$begingroup$
@mina_world: All the text should be actual text. Someone might remember that there was some recent question involving, say, the terms "midpoint" and "collinear"; since "collinear" is trapped in the image, it doesn't get read. (This isn't the best example, but I hope you take my meaning: You want as much of your work as possible to be searchable.) As I mentioned in a comment to another item of yours: screen readers (for the visually impaired) need text. (Granted, MathJax is also problematic for screen readers, but there are occasional words in there. :)
$endgroup$
– Blue
Jan 24 at 16:22
1
$begingroup$
While I'm handing out friendly advice :) ... It would be really helpful to provide more-informative titles to your questions. "Geometry with triangle ABC", "Geometry with circle and triangle", "Geometry with triangle" don't tell the reader what to expect. Moreover, if these questions had come up in a search, the searcher wouldn't be able to tell which questions are actually relevant. (Years from now, when you decide to look back through your own questions for something that could help someone, you won't be able to tell which is which. Try to give questions titles that help "Future You".)
$endgroup$
– Blue
Jan 24 at 16:25
3
$begingroup$
@Blue I will not be satisfied with everything, but I will always do my best. Thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:31
5
5
$begingroup$
In the future, please use MathJax to format your mathematics, because this makes your question more search-engine friendly. Having said that, though, the way you've typeset your question is beautiful and very easy to read.
$endgroup$
– John
Jan 24 at 15:58
$begingroup$
In the future, please use MathJax to format your mathematics, because this makes your question more search-engine friendly. Having said that, though, the way you've typeset your question is beautiful and very easy to read.
$endgroup$
– John
Jan 24 at 15:58
$begingroup$
@John In fact, that's why I always write the problem with MathJax in the first place. (for search).
$endgroup$
– mina_world
Jan 24 at 16:05
$begingroup$
@John In fact, that's why I always write the problem with MathJax in the first place. (for search).
$endgroup$
– mina_world
Jan 24 at 16:05
1
1
$begingroup$
@mina_world: All the text should be actual text. Someone might remember that there was some recent question involving, say, the terms "midpoint" and "collinear"; since "collinear" is trapped in the image, it doesn't get read. (This isn't the best example, but I hope you take my meaning: You want as much of your work as possible to be searchable.) As I mentioned in a comment to another item of yours: screen readers (for the visually impaired) need text. (Granted, MathJax is also problematic for screen readers, but there are occasional words in there. :)
$endgroup$
– Blue
Jan 24 at 16:22
$begingroup$
@mina_world: All the text should be actual text. Someone might remember that there was some recent question involving, say, the terms "midpoint" and "collinear"; since "collinear" is trapped in the image, it doesn't get read. (This isn't the best example, but I hope you take my meaning: You want as much of your work as possible to be searchable.) As I mentioned in a comment to another item of yours: screen readers (for the visually impaired) need text. (Granted, MathJax is also problematic for screen readers, but there are occasional words in there. :)
$endgroup$
– Blue
Jan 24 at 16:22
1
1
$begingroup$
While I'm handing out friendly advice :) ... It would be really helpful to provide more-informative titles to your questions. "Geometry with triangle ABC", "Geometry with circle and triangle", "Geometry with triangle" don't tell the reader what to expect. Moreover, if these questions had come up in a search, the searcher wouldn't be able to tell which questions are actually relevant. (Years from now, when you decide to look back through your own questions for something that could help someone, you won't be able to tell which is which. Try to give questions titles that help "Future You".)
$endgroup$
– Blue
Jan 24 at 16:25
$begingroup$
While I'm handing out friendly advice :) ... It would be really helpful to provide more-informative titles to your questions. "Geometry with triangle ABC", "Geometry with circle and triangle", "Geometry with triangle" don't tell the reader what to expect. Moreover, if these questions had come up in a search, the searcher wouldn't be able to tell which questions are actually relevant. (Years from now, when you decide to look back through your own questions for something that could help someone, you won't be able to tell which is which. Try to give questions titles that help "Future You".)
$endgroup$
– Blue
Jan 24 at 16:25
3
3
$begingroup$
@Blue I will not be satisfied with everything, but I will always do my best. Thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:31
$begingroup$
@Blue I will not be satisfied with everything, but I will always do my best. Thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:31
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
Usually problems like this have one-line solution using Menelaus's theorem. For triangle $ACE$ and line $BDF$:
$$
left|frac{AF}{FC}frac{CB}{BE}frac{ED}{DA}right| = 1,qquad
frac{AF}{FC} = frac{BE}{CB}frac{DA}{ED} = frac12 frac{4}{3} = frac 23\
$$
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
$endgroup$
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
add a comment |
$begingroup$
Draw a paralel to $BF$ through $E$ and let it cuts $AC$ in $P$.
Then by Thales theorem we have $$FP:PC = BE:EC = 1:1$$
Again by Thales theorem we have $$AF:FP= AD:DE = 4:3$$
So, if $FP = 3x$ then $PC =3x$ and $AF = 4x$ so $$ AF:FC = 4x:6x = 2:3$$
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
|
show 4 more comments
$begingroup$
There is a solution using the following
Theorem : If two triangles have their bases on a same line and have a common height, we have :
$$frac{A_1}{A_2} = frac{L_1}{L_2}$$
meaning that the ratio of their areas is the same as the ratio of the lengths of their bases. See for example http://www.madeiracityschools.org/userfiles/376/Classes/17052/Ratio%20of%20Areas-0.pdf.
Take a look at the following picture where the lower case letters indicate areas of the corresponding triangles.
We can write the following linear system where equations (1) to (4) are direct consequences of the upsaid theorem, the last equation (5) being a normalization (we are free to take the area unit we want ; taking $70$ is the result of an "afterthought", in order to provide integer values ) :
$$begin{cases}
(1)&x/z&=&4/3\
(2)&y/v&=&4/3\
(3)&v+z&=&w\
(4)&v+w+y&=&x+z\
(5)&v+w+x+y+z&=&70
end{cases}$$
out of which one gets
$$(v,w,x,y,z) = (6, 21, 20, 8, 15).$$
The looked-for ratio is thus, using (backwards) the above theorem :
$$frac{AF}{FC}=frac{area(AEF)}{area(ECF)}=frac{y+v}{w}=frac{14}{21}=frac{2}{3}.$$
Remark : Had we taken $1$ instead of $70$ in the RHS of equation (5), we would have obtained the same final result, but with unaesthetic fraction computations.
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
$endgroup$
add a comment |
$begingroup$
Let me put together the two articles that answered my question.
answered Jan 25 at 8:23
mina_worldmina_world
1779
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086014%2fgeometry-with-triangle-abc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Usually problems like this have one-line solution using Menelaus's theorem. For triangle $ACE$ and line $BDF$:
$$
left|frac{AF}{FC}frac{CB}{BE}frac{ED}{DA}right| = 1,qquad
frac{AF}{FC} = frac{BE}{CB}frac{DA}{ED} = frac12 frac{4}{3} = frac 23\
$$
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
$endgroup$
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
add a comment |
$begingroup$
Usually problems like this have one-line solution using Menelaus's theorem. For triangle $ACE$ and line $BDF$:
$$
left|frac{AF}{FC}frac{CB}{BE}frac{ED}{DA}right| = 1,qquad
frac{AF}{FC} = frac{BE}{CB}frac{DA}{ED} = frac12 frac{4}{3} = frac 23\
$$
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
$endgroup$
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
add a comment |
$begingroup$
Usually problems like this have one-line solution using Menelaus's theorem. For triangle $ACE$ and line $BDF$:
$$
left|frac{AF}{FC}frac{CB}{BE}frac{ED}{DA}right| = 1,qquad
frac{AF}{FC} = frac{BE}{CB}frac{DA}{ED} = frac12 frac{4}{3} = frac 23\
$$
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
$endgroup$
Usually problems like this have one-line solution using Menelaus's theorem. For triangle $ACE$ and line $BDF$:
$$
left|frac{AF}{FC}frac{CB}{BE}frac{ED}{DA}right| = 1,qquad
frac{AF}{FC} = frac{BE}{CB}frac{DA}{ED} = frac12 frac{4}{3} = frac 23\
$$
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
answered Jan 24 at 15:59
Vasily MitchVasily Mitch
2,3241311
2,3241311
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
add a comment |
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
add a comment |
$begingroup$
Draw a paralel to $BF$ through $E$ and let it cuts $AC$ in $P$.
Then by Thales theorem we have $$FP:PC = BE:EC = 1:1$$
Again by Thales theorem we have $$AF:FP= AD:DE = 4:3$$
So, if $FP = 3x$ then $PC =3x$ and $AF = 4x$ so $$ AF:FC = 4x:6x = 2:3$$
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
|
show 4 more comments
$begingroup$
Draw a paralel to $BF$ through $E$ and let it cuts $AC$ in $P$.
Then by Thales theorem we have $$FP:PC = BE:EC = 1:1$$
Again by Thales theorem we have $$AF:FP= AD:DE = 4:3$$
So, if $FP = 3x$ then $PC =3x$ and $AF = 4x$ so $$ AF:FC = 4x:6x = 2:3$$
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
|
show 4 more comments
$begingroup$
Draw a paralel to $BF$ through $E$ and let it cuts $AC$ in $P$.
Then by Thales theorem we have $$FP:PC = BE:EC = 1:1$$
Again by Thales theorem we have $$AF:FP= AD:DE = 4:3$$
So, if $FP = 3x$ then $PC =3x$ and $AF = 4x$ so $$ AF:FC = 4x:6x = 2:3$$
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
$endgroup$
Draw a paralel to $BF$ through $E$ and let it cuts $AC$ in $P$.
Then by Thales theorem we have $$FP:PC = BE:EC = 1:1$$
Again by Thales theorem we have $$AF:FP= AD:DE = 4:3$$
So, if $FP = 3x$ then $PC =3x$ and $AF = 4x$ so $$ AF:FC = 4x:6x = 2:3$$
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
edited Jan 24 at 21:21
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
answered Jan 24 at 15:55
greedoidgreedoid
42.7k1153105
42.7k1153105
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
|
show 4 more comments
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
Very nice! thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:08
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
In what country do they call it Thales theorem?
$endgroup$
– BPP
Jan 24 at 21:20
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
Grecee, Croatia, Romania, Bulgaria, Serbia, Bosna and Hercegovina, Austria,...
$endgroup$
– greedoid
Jan 24 at 21:22
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
I thought only in France and Switzerland.
$endgroup$
– BPP
Jan 24 at 21:32
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
$begingroup$
Well, I didn't know that ...
$endgroup$
– greedoid
Jan 24 at 21:33
|
show 4 more comments
$begingroup$
There is a solution using the following
Theorem : If two triangles have their bases on a same line and have a common height, we have :
$$frac{A_1}{A_2} = frac{L_1}{L_2}$$
meaning that the ratio of their areas is the same as the ratio of the lengths of their bases. See for example http://www.madeiracityschools.org/userfiles/376/Classes/17052/Ratio%20of%20Areas-0.pdf.
Take a look at the following picture where the lower case letters indicate areas of the corresponding triangles.
We can write the following linear system where equations (1) to (4) are direct consequences of the upsaid theorem, the last equation (5) being a normalization (we are free to take the area unit we want ; taking $70$ is the result of an "afterthought", in order to provide integer values ) :
$$begin{cases}
(1)&x/z&=&4/3\
(2)&y/v&=&4/3\
(3)&v+z&=&w\
(4)&v+w+y&=&x+z\
(5)&v+w+x+y+z&=&70
end{cases}$$
out of which one gets
$$(v,w,x,y,z) = (6, 21, 20, 8, 15).$$
The looked-for ratio is thus, using (backwards) the above theorem :
$$frac{AF}{FC}=frac{area(AEF)}{area(ECF)}=frac{y+v}{w}=frac{14}{21}=frac{2}{3}.$$
Remark : Had we taken $1$ instead of $70$ in the RHS of equation (5), we would have obtained the same final result, but with unaesthetic fraction computations.
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
$endgroup$
add a comment |
$begingroup$
There is a solution using the following
Theorem : If two triangles have their bases on a same line and have a common height, we have :
$$frac{A_1}{A_2} = frac{L_1}{L_2}$$
meaning that the ratio of their areas is the same as the ratio of the lengths of their bases. See for example http://www.madeiracityschools.org/userfiles/376/Classes/17052/Ratio%20of%20Areas-0.pdf.
Take a look at the following picture where the lower case letters indicate areas of the corresponding triangles.
We can write the following linear system where equations (1) to (4) are direct consequences of the upsaid theorem, the last equation (5) being a normalization (we are free to take the area unit we want ; taking $70$ is the result of an "afterthought", in order to provide integer values ) :
$$begin{cases}
(1)&x/z&=&4/3\
(2)&y/v&=&4/3\
(3)&v+z&=&w\
(4)&v+w+y&=&x+z\
(5)&v+w+x+y+z&=&70
end{cases}$$
out of which one gets
$$(v,w,x,y,z) = (6, 21, 20, 8, 15).$$
The looked-for ratio is thus, using (backwards) the above theorem :
$$frac{AF}{FC}=frac{area(AEF)}{area(ECF)}=frac{y+v}{w}=frac{14}{21}=frac{2}{3}.$$
Remark : Had we taken $1$ instead of $70$ in the RHS of equation (5), we would have obtained the same final result, but with unaesthetic fraction computations.
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
$endgroup$
add a comment |
$begingroup$
There is a solution using the following
Theorem : If two triangles have their bases on a same line and have a common height, we have :
$$frac{A_1}{A_2} = frac{L_1}{L_2}$$
meaning that the ratio of their areas is the same as the ratio of the lengths of their bases. See for example http://www.madeiracityschools.org/userfiles/376/Classes/17052/Ratio%20of%20Areas-0.pdf.
Take a look at the following picture where the lower case letters indicate areas of the corresponding triangles.
We can write the following linear system where equations (1) to (4) are direct consequences of the upsaid theorem, the last equation (5) being a normalization (we are free to take the area unit we want ; taking $70$ is the result of an "afterthought", in order to provide integer values ) :
$$begin{cases}
(1)&x/z&=&4/3\
(2)&y/v&=&4/3\
(3)&v+z&=&w\
(4)&v+w+y&=&x+z\
(5)&v+w+x+y+z&=&70
end{cases}$$
out of which one gets
$$(v,w,x,y,z) = (6, 21, 20, 8, 15).$$
The looked-for ratio is thus, using (backwards) the above theorem :
$$frac{AF}{FC}=frac{area(AEF)}{area(ECF)}=frac{y+v}{w}=frac{14}{21}=frac{2}{3}.$$
Remark : Had we taken $1$ instead of $70$ in the RHS of equation (5), we would have obtained the same final result, but with unaesthetic fraction computations.
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
$endgroup$
There is a solution using the following
Theorem : If two triangles have their bases on a same line and have a common height, we have :
$$frac{A_1}{A_2} = frac{L_1}{L_2}$$
meaning that the ratio of their areas is the same as the ratio of the lengths of their bases. See for example http://www.madeiracityschools.org/userfiles/376/Classes/17052/Ratio%20of%20Areas-0.pdf.
Take a look at the following picture where the lower case letters indicate areas of the corresponding triangles.
We can write the following linear system where equations (1) to (4) are direct consequences of the upsaid theorem, the last equation (5) being a normalization (we are free to take the area unit we want ; taking $70$ is the result of an "afterthought", in order to provide integer values ) :
$$begin{cases}
(1)&x/z&=&4/3\
(2)&y/v&=&4/3\
(3)&v+z&=&w\
(4)&v+w+y&=&x+z\
(5)&v+w+x+y+z&=&70
end{cases}$$
out of which one gets
$$(v,w,x,y,z) = (6, 21, 20, 8, 15).$$
The looked-for ratio is thus, using (backwards) the above theorem :
$$frac{AF}{FC}=frac{area(AEF)}{area(ECF)}=frac{y+v}{w}=frac{14}{21}=frac{2}{3}.$$
Remark : Had we taken $1$ instead of $70$ in the RHS of equation (5), we would have obtained the same final result, but with unaesthetic fraction computations.
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
edited Jan 25 at 23:37
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
answered Jan 25 at 10:13
Jean MarieJean Marie
29.9k42051
29.9k42051
add a comment |
add a comment |
$begingroup$
Let me put together the two articles that answered my question.
answered Jan 25 at 8:23
mina_worldmina_world
1779
$endgroup$
add a comment |
$begingroup$
Let me put together the two articles that answered my question.
answered Jan 25 at 8:23
mina_worldmina_world
1779
$endgroup$
add a comment |
$begingroup$
Let me put together the two articles that answered my question.
answered Jan 25 at 8:23
mina_worldmina_world
1779
$endgroup$
Let me put together the two articles that answered my question.
answered Jan 25 at 8:23
mina_worldmina_world
1779
answered Jan 25 at 8:23
mina_worldmina_world
1779
answered Jan 25 at 8:23
mina_worldmina_world
1779
answered Jan 25 at 8:23
mina_worldmina_world
1779
1779
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3086014%2fgeometry-with-triangle-abc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
5
$begingroup$
In the future, please use MathJax to format your mathematics, because this makes your question more search-engine friendly. Having said that, though, the way you've typeset your question is beautiful and very easy to read.
$endgroup$
– John
Jan 24 at 15:58
$begingroup$
@John In fact, that's why I always write the problem with MathJax in the first place. (for search).
$endgroup$
– mina_world
Jan 24 at 16:05
1
$begingroup$
@mina_world: All the text should be actual text. Someone might remember that there was some recent question involving, say, the terms "midpoint" and "collinear"; since "collinear" is trapped in the image, it doesn't get read. (This isn't the best example, but I hope you take my meaning: You want as much of your work as possible to be searchable.) As I mentioned in a comment to another item of yours: screen readers (for the visually impaired) need text. (Granted, MathJax is also problematic for screen readers, but there are occasional words in there. :)
$endgroup$
– Blue
Jan 24 at 16:22
1
$begingroup$
While I'm handing out friendly advice :) ... It would be really helpful to provide more-informative titles to your questions. "Geometry with triangle ABC", "Geometry with circle and triangle", "Geometry with triangle" don't tell the reader what to expect. Moreover, if these questions had come up in a search, the searcher wouldn't be able to tell which questions are actually relevant. (Years from now, when you decide to look back through your own questions for something that could help someone, you won't be able to tell which is which. Try to give questions titles that help "Future You".)
$endgroup$
– Blue
Jan 24 at 16:25
3
$begingroup$
@Blue I will not be satisfied with everything, but I will always do my best. Thank you very much.
$endgroup$
– mina_world
Jan 24 at 16:31