Quadrilaterals with equal sides Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)On QuadrilateralsIn this isosceles right angled triangle, prove that $angle DAE = 45^circ$A “Paradoxist Geometry”Congruence of quadrilaterals given the sidesUsing angle chasing to find $angleDBF$Proof of equal angles in a quadrilateral.Points where no lines intersect both sides of an angle in the Poincaré's disk model.In the figure: prove that $overline AM perp overline DE$ if $M$ is the middle point on $overline BC$.Calculate angle EZB in the below drawing.Locus of a point on triangle side
How do I find out the mythology and history of my Fortress?
Subalgebra of a group algebra
How often does castling occur in grandmaster games?
Belief In God or Knowledge Of God. Which is better?
How to improve on this Stylesheet Manipulation for Message Styling
Why do early math courses focus on the cross sections of a cone and not on other 3D objects?
What is the appropriate index architecture when forced to implement IsDeleted (soft deletes)?
Putting class ranking in CV, but against dept guidelines
Can a new player join a group only when a new campaign starts?
Effects on objects due to a brief relocation of massive amounts of mass
Time to Settle Down!
Strange behavior of Object.defineProperty() in JavaScript
Triggering an ultrasonic sensor
Exposing GRASS GIS add-on in QGIS Processing framework?
Should I use a zero-interest credit card for a large one-time purchase?
Sum letters are not two different
Would it be possible to dictate a bech32 address as a list of English words?
Why weren't discrete x86 CPUs ever used in game hardware?
What is the meaning of 'breadth' in breadth first search?
Is CEO the "profession" with the most psychopaths?
Is there hard evidence that the grant peer review system performs significantly better than random?
Product of Mrówka space and one point compactification discrete space.
How to run automated tests after each commit?
Parallel Computing Problem
Quadrilaterals with equal sides
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)On QuadrilateralsIn this isosceles right angled triangle, prove that $angle DAE = 45^circ$A “Paradoxist Geometry”Congruence of quadrilaterals given the sidesUsing angle chasing to find $angleDBF$Proof of equal angles in a quadrilateral.Points where no lines intersect both sides of an angle in the Poincaré's disk model.In the figure: prove that $overline AM perp overline DE$ if $M$ is the middle point on $overline BC$.Calculate angle EZB in the below drawing.Locus of a point on triangle side
$begingroup$
$AC = BD$
$EC = ED$
$AF = FB$
Angle CAF = 70 deg
Angle DBF = 60 deg
We are looking for angle EFA.
I have found through Geogebra that the required angle is 85 deg.
Any ideas how to prove it? I am not so familiar with Geometry :(
geometry
asked 4 hours ago
SamuelSamuel
502412
$endgroup$
add a comment |
$begingroup$
$AC = BD$
$EC = ED$
$AF = FB$
Angle CAF = 70 deg
Angle DBF = 60 deg
We are looking for angle EFA.
I have found through Geogebra that the required angle is 85 deg.
Any ideas how to prove it? I am not so familiar with Geometry :(
geometry
asked 4 hours ago
SamuelSamuel
502412
$endgroup$
add a comment |
$begingroup$
$AC = BD$
$EC = ED$
$AF = FB$
Angle CAF = 70 deg
Angle DBF = 60 deg
We are looking for angle EFA.
I have found through Geogebra that the required angle is 85 deg.
Any ideas how to prove it? I am not so familiar with Geometry :(
geometry
asked 4 hours ago
SamuelSamuel
502412
$endgroup$
$AC = BD$
$EC = ED$
$AF = FB$
Angle CAF = 70 deg
Angle DBF = 60 deg
We are looking for angle EFA.
I have found through Geogebra that the required angle is 85 deg.
Any ideas how to prove it? I am not so familiar with Geometry :(
geometry
geometry
asked 4 hours ago
SamuelSamuel
502412
asked 4 hours ago
SamuelSamuel
502412
asked 4 hours ago
SamuelSamuel
502412
asked 4 hours ago
SamuelSamuel
502412
asked 4 hours ago
SamuelSamuel
502412
502412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the following triangle:-
Let $JM = a, JN = b $ . In this particular $triangle$, $MK=NL =$ say $x$.
Draw the angle bisector of $angle J$ , $JO$.
WLOG $a<b$.
Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.
Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$
Now , locate a point $J'$ along $JL$ , such that $JJ'=fracb-a2$ . While this may seem arbitrary , things will become clear soon.
Draw $J'Q$ parallel to $JO$ .
$J'N=JN-JJ'=fraca+b2$.
Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frack_1(a+b)2$
This implies that $S$ is the midpoint of $MN$ !
Similarly , we find $QL$ to equal $k_2(fraca+b2+x)$ , proving that $Q$ is the midpoint of $KL$ .
Recall that by construction, $J'Q$ is parallel to $JO$.
Thus , we have discovered the fact , that :-
The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.
Your problem is now trivial .
In your case , $angle MKL=70 , angle KLN =60 $
$therefore angle KJL = 50 implies angle RJN = angle QJ'N = 25$
External angle $J'QK = 25+60 = boxed85 $
answered 1 hour ago
SinπSinπ
71011
$endgroup$
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
add a comment |
Your Answer
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%2f3193339%2fquadrilaterals-with-equal-sides%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the following triangle:-
Let $JM = a, JN = b $ . In this particular $triangle$, $MK=NL =$ say $x$.
Draw the angle bisector of $angle J$ , $JO$.
WLOG $a<b$.
Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.
Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$
Now , locate a point $J'$ along $JL$ , such that $JJ'=fracb-a2$ . While this may seem arbitrary , things will become clear soon.
Draw $J'Q$ parallel to $JO$ .
$J'N=JN-JJ'=fraca+b2$.
Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frack_1(a+b)2$
This implies that $S$ is the midpoint of $MN$ !
Similarly , we find $QL$ to equal $k_2(fraca+b2+x)$ , proving that $Q$ is the midpoint of $KL$ .
Recall that by construction, $J'Q$ is parallel to $JO$.
Thus , we have discovered the fact , that :-
The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.
Your problem is now trivial .
In your case , $angle MKL=70 , angle KLN =60 $
$therefore angle KJL = 50 implies angle RJN = angle QJ'N = 25$
External angle $J'QK = 25+60 = boxed85 $
answered 1 hour ago
SinπSinπ
71011
$endgroup$
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
add a comment |
$begingroup$
Consider the following triangle:-
Let $JM = a, JN = b $ . In this particular $triangle$, $MK=NL =$ say $x$.
Draw the angle bisector of $angle J$ , $JO$.
WLOG $a<b$.
Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.
Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$
Now , locate a point $J'$ along $JL$ , such that $JJ'=fracb-a2$ . While this may seem arbitrary , things will become clear soon.
Draw $J'Q$ parallel to $JO$ .
$J'N=JN-JJ'=fraca+b2$.
Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frack_1(a+b)2$
This implies that $S$ is the midpoint of $MN$ !
Similarly , we find $QL$ to equal $k_2(fraca+b2+x)$ , proving that $Q$ is the midpoint of $KL$ .
Recall that by construction, $J'Q$ is parallel to $JO$.
Thus , we have discovered the fact , that :-
The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.
Your problem is now trivial .
In your case , $angle MKL=70 , angle KLN =60 $
$therefore angle KJL = 50 implies angle RJN = angle QJ'N = 25$
External angle $J'QK = 25+60 = boxed85 $
answered 1 hour ago
SinπSinπ
71011
$endgroup$
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
add a comment |
$begingroup$
Consider the following triangle:-
Let $JM = a, JN = b $ . In this particular $triangle$, $MK=NL =$ say $x$.
Draw the angle bisector of $angle J$ , $JO$.
WLOG $a<b$.
Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.
Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$
Now , locate a point $J'$ along $JL$ , such that $JJ'=fracb-a2$ . While this may seem arbitrary , things will become clear soon.
Draw $J'Q$ parallel to $JO$ .
$J'N=JN-JJ'=fraca+b2$.
Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frack_1(a+b)2$
This implies that $S$ is the midpoint of $MN$ !
Similarly , we find $QL$ to equal $k_2(fraca+b2+x)$ , proving that $Q$ is the midpoint of $KL$ .
Recall that by construction, $J'Q$ is parallel to $JO$.
Thus , we have discovered the fact , that :-
The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.
Your problem is now trivial .
In your case , $angle MKL=70 , angle KLN =60 $
$therefore angle KJL = 50 implies angle RJN = angle QJ'N = 25$
External angle $J'QK = 25+60 = boxed85 $
answered 1 hour ago
SinπSinπ
71011
$endgroup$
Consider the following triangle:-
Let $JM = a, JN = b $ . In this particular $triangle$, $MK=NL =$ say $x$.
Draw the angle bisector of $angle J$ , $JO$.
WLOG $a<b$.
Then , by the internal angle bisector theorem , $MR = k_1a , RN = k_1b , KO= k_2(a+x) , OL = k_2(b+x) $.
Obviously , $MN=k_1(a+b) $ and $KL =k_2(a+b+2x)$
Now , locate a point $J'$ along $JL$ , such that $JJ'=fracb-a2$ . While this may seem arbitrary , things will become clear soon.
Draw $J'Q$ parallel to $JO$ .
$J'N=JN-JJ'=fraca+b2$.
Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frack_1(a+b)2$
This implies that $S$ is the midpoint of $MN$ !
Similarly , we find $QL$ to equal $k_2(fraca+b2+x)$ , proving that $Q$ is the midpoint of $KL$ .
Recall that by construction, $J'Q$ is parallel to $JO$.
Thus , we have discovered the fact , that :-
The line joining the midpoints of the opposite sides of a quadrilateral ,when its other sides are equal , is parallel to the angle bisector of the angle formed by extending the other two sides.
Your problem is now trivial .
In your case , $angle MKL=70 , angle KLN =60 $
$therefore angle KJL = 50 implies angle RJN = angle QJ'N = 25$
External angle $J'QK = 25+60 = boxed85 $
answered 1 hour ago
SinπSinπ
71011
edited 54 mins ago
answered 1 hour ago
SinπSinπ
71011
answered 1 hour ago
SinπSinπ
71011
answered 1 hour ago
SinπSinπ
71011
71011
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
add a comment |
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
Hope you retained OP's labels of intersection points.
$endgroup$
– Narasimham
41 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire, but it would take a while.
$endgroup$
– Sinπ
36 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
$begingroup$
@Sinπ thank you very much!! No need to change any labels; everything is very clear!!
$endgroup$
– Samuel
4 mins ago
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%2f3193339%2fquadrilaterals-with-equal-sides%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
