Sfrgttk

How does the secondary effect of the Heat Metal spell interact with a creature resistant/immune to fire damage?

Is there hard evidence that the grant peer review system performs significantly better than random?

Is there any word for a place full of confusion?

"Lost his faith in humanity in the trenches of Verdun" — last line of an SF story

How to draw/optimize this graph with tikz

Take 2! Is this homebrew Lady of Pain warlock patron balanced?

Amount of permutations on an NxNxN Rubik's Cube

Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?

What does it mean that physics no longer uses mechanical models to describe phenomena?

How much damage would a cupful of neutron star matter do to the Earth?

How could we fake a moon landing now?

How would a mousetrap for use in space work?

How many serial ports are on the Pi 3?

Converted a Scalar function to a TVF function for parallel execution-Still running in Serial mode

How to dry out epoxy resin faster than usual?

A term for a woman complaining about things/begging in a cute/childish way

What would you call this weird metallic apparatus that allows you to lift people?

Exposing GRASS GIS add-on in QGIS Processing framework?

AppleTVs create a chatty alternate WiFi network

Can a new player join a group only when a new campaign starts?

Crossing US/Canada Border for less than 24 hours

Why weren't discrete x86 CPUs ever used in game hardware?

Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?

Can you explain what "processes and tools" means in the first Agile principle?

Quadrilaterals with equal sides

10

6

$begingroup$

$AC = BD$

$EC = ED$

$AF = FB$

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

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

495412

$endgroup$

add a comment | 

10

6

$begingroup$

$AC = BD$

$EC = ED$

$AF = FB$

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

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

495412

$endgroup$

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

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

495412

$endgroup$

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

495412

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

495412

asked 4 hours ago

SamuelSamuel

495412

asked 4 hours ago

SamuelSamuel

495412

1 Answer
1

active

oldest

votes

3

$begingroup$

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 $JJ'=frac{b-a}{2}$ . While this may seem arbitrary , things will become clear soon.

Draw $J'Q$ parallel to $JO$ .

$J'N=JN-JJ'=frac{a+b}{2}$.

Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frac{k_1(a+b)}{2}$

This implies that $S$ is the midpoint of $MN$ !

Similarly , we find $QL$ to equal $k_2(frac{a+b}{2}+x)$ , 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 = boxed{85} $

share|cite|improve this answer

edited 17 mins ago

answered 28 mins ago

SinπSinπ

67511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope you retained OP's labels of intersection points.
  $endgroup$
  – Narasimham
  4 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire.
  $endgroup$
  – Sinπ
  just now
  

  
  
  
  

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
});


}
});

draft saved

draft discarded

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

Name

Email

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

Name

Email

Required, but never shown

3

$begingroup$

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 $JJ'=frac{b-a}{2}$ . While this may seem arbitrary , things will become clear soon.

Draw $J'Q$ parallel to $JO$ .

$J'N=JN-JJ'=frac{a+b}{2}$.

Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frac{k_1(a+b)}{2}$

This implies that $S$ is the midpoint of $MN$ !

Similarly , we find $QL$ to equal $k_2(frac{a+b}{2}+x)$ , 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 = boxed{85} $

share|cite|improve this answer

edited 17 mins ago

answered 28 mins ago

SinπSinπ

67511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope you retained OP's labels of intersection points.
  $endgroup$
  – Narasimham
  4 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire.
  $endgroup$
  – Sinπ
  just now
  

  
  
  
  

add a comment | 

3

$begingroup$

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 $JJ'=frac{b-a}{2}$ . While this may seem arbitrary , things will become clear soon.

Draw $J'Q$ parallel to $JO$ .

$J'N=JN-JJ'=frac{a+b}{2}$.

Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frac{k_1(a+b)}{2}$

This implies that $S$ is the midpoint of $MN$ !

Similarly , we find $QL$ to equal $k_2(frac{a+b}{2}+x)$ , 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 = boxed{85} $

share|cite|improve this answer

edited 17 mins ago

answered 28 mins ago

SinπSinπ

67511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope you retained OP's labels of intersection points.
  $endgroup$
  – Narasimham
  4 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Narasimham No , I'm sorry , I haven't ;) . I can change all the labels , and the image , if you desire.
  $endgroup$
  – Sinπ
  just now
  

  
  
  
  

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 $JJ'=frac{b-a}{2}$ . While this may seem arbitrary , things will become clear soon.

Draw $J'Q$ parallel to $JO$ .

$J'N=JN-JJ'=frac{a+b}{2}$.

Using similarity in $triangle $s $JRN$ and $J'SN$ , $SN$ = $frac{k_1(a+b)}{2}$

This implies that $S$ is the midpoint of $MN$ !

Similarly , we find $QL$ to equal $k_2(frac{a+b}{2}+x)$ , 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 = boxed{85} $

share|cite|improve this answer

edited 17 mins ago

answered 28 mins ago

SinπSinπ

67511

$endgroup$

share|cite|improve this answer

edited 17 mins ago

answered 28 mins ago

SinπSinπ

67511

answered 28 mins ago

SinπSinπ

67511

answered 28 mins ago

SinπSinπ

67511