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Quadrilaterals with equal sides

11

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

500412

$endgroup$

add a comment | 

11

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

500412

$endgroup$

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

500412

$endgroup$

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

500412

share|cite|improve this question

asked 4 hours ago

SamuelSamuel

500412

asked 4 hours ago

SamuelSamuel

500412

asked 4 hours ago

SamuelSamuel

500412

1 Answer
1

active

oldest

votes

4

$begingroup$

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 $

share|cite|improve this answer

edited 42 mins ago

answered 53 mins ago

SinπSinπ

68511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope you retained OP's labels of intersection points.
  $endgroup$
  – Narasimham
  28 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
  24 mins ago
  
  

  
  
  
  

add a comment | 

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4

$begingroup$

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 $

share|cite|improve this answer

edited 42 mins ago

answered 53 mins ago

SinπSinπ

68511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope you retained OP's labels of intersection points.
  $endgroup$
  – Narasimham
  28 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
  24 mins ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

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 $

share|cite|improve this answer

edited 42 mins ago

answered 53 mins ago

SinπSinπ

68511

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope you retained OP's labels of intersection points.
  $endgroup$
  – Narasimham
  28 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
  24 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 $

share|cite|improve this answer

edited 42 mins ago

answered 53 mins ago

SinπSinπ

68511

$endgroup$

share|cite|improve this answer

edited 42 mins ago

answered 53 mins ago

SinπSinπ

68511

answered 53 mins ago

SinπSinπ

68511

answered 53 mins ago

SinπSinπ

68511