Finding angles with bisectors

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Let ABC be a triangle with ∠A=90 and ∠B =20 . Let E and F be points on
AC and AB respectively such that ∠ABE = 10 and ∠ACF = 30 . Determine
∠CFE .

The answer is 20 but it doesn't show in my solution:

Let the intersection of EB and FC be H. BH is an angle bisector.

$fracBFBC = fracFHCH$

Let angle CFE be A. then EHF is 50-A (angle chasing).

Using Sine Law on triangles EFH and EHC and BFC, we have

$fracEHFH = fracsin(a)sin(50-a)$

$fracEHCH =fracsin(30)sin(100)$

$fracBFBC=fracsin(40)sin(100)$

using the 4 equations, I got
$sin(100)sin(a)sin(40)= sin(30)sin(120)sin(50-a)$
but obviously A is not 20, can someone spot my mistake, I cant seem to pinpoint it.

geometry

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asked Jul 23 at 7:20

SuperMage1

661210

add a comment | 

up vote
0
down vote

favorite

.

Let ABC be a triangle with ∠A=90 and ∠B =20 . Let E and F be points on
AC and AB respectively such that ∠ABE = 10 and ∠ACF = 30 . Determine
∠CFE .

The answer is 20 but it doesn't show in my solution:

Let the intersection of EB and FC be H. BH is an angle bisector.

$fracBFBC = fracFHCH$

Let angle CFE be A. then EHF is 50-A (angle chasing).

Using Sine Law on triangles EFH and EHC and BFC, we have

$fracEHFH = fracsin(a)sin(50-a)$

$fracEHCH =fracsin(30)sin(100)$

$fracBFBC=fracsin(40)sin(100)$

using the 4 equations, I got
$sin(100)sin(a)sin(40)= sin(30)sin(120)sin(50-a)$
but obviously A is not 20, can someone spot my mistake, I cant seem to pinpoint it.

geometry

share|cite|improve this question

asked Jul 23 at 7:20

SuperMage1

661210

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

.

Let ABC be a triangle with ∠A=90 and ∠B =20 . Let E and F be points on
AC and AB respectively such that ∠ABE = 10 and ∠ACF = 30 . Determine
∠CFE .

The answer is 20 but it doesn't show in my solution:

Let the intersection of EB and FC be H. BH is an angle bisector.

$fracBFBC = fracFHCH$

Let angle CFE be A. then EHF is 50-A (angle chasing).

Using Sine Law on triangles EFH and EHC and BFC, we have

$fracEHFH = fracsin(a)sin(50-a)$

$fracEHCH =fracsin(30)sin(100)$

$fracBFBC=fracsin(40)sin(100)$

using the 4 equations, I got
$sin(100)sin(a)sin(40)= sin(30)sin(120)sin(50-a)$
but obviously A is not 20, can someone spot my mistake, I cant seem to pinpoint it.

geometry

share|cite|improve this question

asked Jul 23 at 7:20

SuperMage1

661210

.

Let ABC be a triangle with ∠A=90 and ∠B =20 . Let E and F be points on
AC and AB respectively such that ∠ABE = 10 and ∠ACF = 30 . Determine
∠CFE .

The answer is 20 but it doesn't show in my solution:

Let the intersection of EB and FC be H. BH is an angle bisector.

$fracBFBC = fracFHCH$

Let angle CFE be A. then EHF is 50-A (angle chasing).

Using Sine Law on triangles EFH and EHC and BFC, we have

$fracEHFH = fracsin(a)sin(50-a)$

$fracEHCH =fracsin(30)sin(100)$

$fracBFBC=fracsin(40)sin(100)$

using the 4 equations, I got
$sin(100)sin(a)sin(40)= sin(30)sin(120)sin(50-a)$
but obviously A is not 20, can someone spot my mistake, I cant seem to pinpoint it.

geometry

share|cite|improve this question

asked Jul 23 at 7:20

SuperMage1

661210

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 7:20

SuperMage1

661210

asked Jul 23 at 7:20

SuperMage1

661210

asked Jul 23 at 7:20

SuperMage1

661210

661210

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add a comment | 

1 Answer
1

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accepted

Note that $FC = 2AF$. Let $D$ be the midpoint of $BC$ and let $G$ be the point on $AB$ such that $GD$ is perpendicular to $BC$. Then triangles $ABC$ and $DBG$ are similar, so that $dfracBDBG=dfracBABC$. Then by symmetry we have $angle GCB=angle GBC=20^circ$, so that $angle GCF=20^circ$ also. Hence $CG$ bisects $angle BCF$ so that $dfracFCFG=dfracBCBG$. Since $BE$ bisects $angle ABC$, $dfracBABC=dfracAECE$. Now
$$fracAFFG=fracdfrac12FCFG=fracdfrac12 BC BG=fracBDBG=fracBABC=fracAEEC$$

It follows that $CG$ is parallel to $EF$, so that $angle CFE=angle GCF=20^circ$

share|cite|improve this answer

edited Jul 23 at 11:36

answered Jul 23 at 11:15

Key Flex

4,258423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  on the third line , GCB =GCB?
  – SuperMage1
  Jul 23 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited, it should be $angle GCB=angle GBC$
  – Key Flex
  Jul 23 at 11:37
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Note that $FC = 2AF$. Let $D$ be the midpoint of $BC$ and let $G$ be the point on $AB$ such that $GD$ is perpendicular to $BC$. Then triangles $ABC$ and $DBG$ are similar, so that $dfracBDBG=dfracBABC$. Then by symmetry we have $angle GCB=angle GBC=20^circ$, so that $angle GCF=20^circ$ also. Hence $CG$ bisects $angle BCF$ so that $dfracFCFG=dfracBCBG$. Since $BE$ bisects $angle ABC$, $dfracBABC=dfracAECE$. Now
$$fracAFFG=fracdfrac12FCFG=fracdfrac12 BC BG=fracBDBG=fracBABC=fracAEEC$$

It follows that $CG$ is parallel to $EF$, so that $angle CFE=angle GCF=20^circ$

share|cite|improve this answer

edited Jul 23 at 11:36

answered Jul 23 at 11:15

Key Flex

4,258423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  on the third line , GCB =GCB?
  – SuperMage1
  Jul 23 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited, it should be $angle GCB=angle GBC$
  – Key Flex
  Jul 23 at 11:37
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Note that $FC = 2AF$. Let $D$ be the midpoint of $BC$ and let $G$ be the point on $AB$ such that $GD$ is perpendicular to $BC$. Then triangles $ABC$ and $DBG$ are similar, so that $dfracBDBG=dfracBABC$. Then by symmetry we have $angle GCB=angle GBC=20^circ$, so that $angle GCF=20^circ$ also. Hence $CG$ bisects $angle BCF$ so that $dfracFCFG=dfracBCBG$. Since $BE$ bisects $angle ABC$, $dfracBABC=dfracAECE$. Now
$$fracAFFG=fracdfrac12FCFG=fracdfrac12 BC BG=fracBDBG=fracBABC=fracAEEC$$

It follows that $CG$ is parallel to $EF$, so that $angle CFE=angle GCF=20^circ$

share|cite|improve this answer

edited Jul 23 at 11:36

answered Jul 23 at 11:15

Key Flex

4,258423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  on the third line , GCB =GCB?
  – SuperMage1
  Jul 23 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited, it should be $angle GCB=angle GBC$
  – Key Flex
  Jul 23 at 11:37
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Note that $FC = 2AF$. Let $D$ be the midpoint of $BC$ and let $G$ be the point on $AB$ such that $GD$ is perpendicular to $BC$. Then triangles $ABC$ and $DBG$ are similar, so that $dfracBDBG=dfracBABC$. Then by symmetry we have $angle GCB=angle GBC=20^circ$, so that $angle GCF=20^circ$ also. Hence $CG$ bisects $angle BCF$ so that $dfracFCFG=dfracBCBG$. Since $BE$ bisects $angle ABC$, $dfracBABC=dfracAECE$. Now
$$fracAFFG=fracdfrac12FCFG=fracdfrac12 BC BG=fracBDBG=fracBABC=fracAEEC$$

It follows that $CG$ is parallel to $EF$, so that $angle CFE=angle GCF=20^circ$

share|cite|improve this answer

edited Jul 23 at 11:36

answered Jul 23 at 11:15

Key Flex

4,258423

Note that $FC = 2AF$. Let $D$ be the midpoint of $BC$ and let $G$ be the point on $AB$ such that $GD$ is perpendicular to $BC$. Then triangles $ABC$ and $DBG$ are similar, so that $dfracBDBG=dfracBABC$. Then by symmetry we have $angle GCB=angle GBC=20^circ$, so that $angle GCF=20^circ$ also. Hence $CG$ bisects $angle BCF$ so that $dfracFCFG=dfracBCBG$. Since $BE$ bisects $angle ABC$, $dfracBABC=dfracAECE$. Now
$$fracAFFG=fracdfrac12FCFG=fracdfrac12 BC BG=fracBDBG=fracBABC=fracAEEC$$

It follows that $CG$ is parallel to $EF$, so that $angle CFE=angle GCF=20^circ$

share|cite|improve this answer

edited Jul 23 at 11:36

answered Jul 23 at 11:15

Key Flex

4,258423

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 11:36

edited Jul 23 at 11:36

edited Jul 23 at 11:36

answered Jul 23 at 11:15

Key Flex

4,258423

answered Jul 23 at 11:15

Key Flex

4,258423

answered Jul 23 at 11:15

Key Flex

4,258423

4,258423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  on the third line , GCB =GCB?
  – SuperMage1
  Jul 23 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited, it should be $angle GCB=angle GBC$
  – Key Flex
  Jul 23 at 11:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  on the third line , GCB =GCB?
  – SuperMage1
  Jul 23 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I edited, it should be $angle GCB=angle GBC$
  – Key Flex
  Jul 23 at 11:37
  

  
  
  
  

on the third line , GCB =GCB?
– SuperMage1
Jul 23 at 11:35

on the third line , GCB =GCB?
– SuperMage1
Jul 23 at 11:35

I edited, it should be $angle GCB=angle GBC$
– Key Flex
Jul 23 at 11:37

I edited, it should be $angle GCB=angle GBC$
– Key Flex
Jul 23 at 11:37

add a comment | 

 

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