Angle bisector in triangle, quick question: $|AE| = frac{bc}{a+c}$

2

$begingroup$

Triangle $ABC$; $AB=c, BC=a, AC=b$; angle bisector of angle $(c, a)$ cuts $AC$ in point $E$.

Why is the following true?
$$|AE| = frac{bc}{a+c}$$
Where does that come from?

geometry trigonometry triangles

share|cite|improve this question

edited Feb 2 at 18:10

Shaun

10.5k113687

asked Feb 2 at 17:59

PeroPero

1497

$endgroup$

add a comment | 

2

$begingroup$

Triangle $ABC$; $AB=c, BC=a, AC=b$; angle bisector of angle $(c, a)$ cuts $AC$ in point $E$.

Why is the following true?
$$|AE| = frac{bc}{a+c}$$
Where does that come from?

geometry trigonometry triangles

share|cite|improve this question

edited Feb 2 at 18:10

Shaun

10.5k113687

asked Feb 2 at 17:59

PeroPero

1497

$endgroup$

add a comment | 

2

2

2

1

$begingroup$

Triangle $ABC$; $AB=c, BC=a, AC=b$; angle bisector of angle $(c, a)$ cuts $AC$ in point $E$.

Why is the following true?
$$|AE| = frac{bc}{a+c}$$
Where does that come from?

geometry trigonometry triangles

share|cite|improve this question

edited Feb 2 at 18:10

Shaun

10.5k113687

asked Feb 2 at 17:59

PeroPero

1497

$endgroup$

Triangle $ABC$; $AB=c, BC=a, AC=b$; angle bisector of angle $(c, a)$ cuts $AC$ in point $E$.

Why is the following true?
$$|AE| = frac{bc}{a+c}$$
Where does that come from?

geometry trigonometry triangles

geometry trigonometry triangles

share|cite|improve this question

edited Feb 2 at 18:10

Shaun

10.5k113687

asked Feb 2 at 17:59

PeroPero

1497

share|cite|improve this question

edited Feb 2 at 18:10

Shaun

10.5k113687

asked Feb 2 at 17:59

PeroPero

1497

share|cite|improve this question

share|cite|improve this question

edited Feb 2 at 18:10

Shaun

10.5k113687

edited Feb 2 at 18:10

Shaun

10.5k113687

edited Feb 2 at 18:10

Shaun

10.5k113687

10.5k113687

asked Feb 2 at 17:59

PeroPero

1497

asked Feb 2 at 17:59

PeroPero

1497

asked Feb 2 at 17:59

PeroPero

1497

1497

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

1

$begingroup$

Hint: We get $$frac{AE}{EC}=frac{c}{a}$$
And $$EC=b-AE$$ then we get $$frac{b-AE}{AE}=frac{c}{a}$$

share|cite|improve this answer

edited Feb 2 at 18:25

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you mean $frac ca$ ? But I still can't get to what I asked from here.
  $endgroup$
  – Pero
  Feb 2 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the angle bisector theorem
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
  $endgroup$
  – Pero
  Feb 2 at 18:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for that typo!
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:10
  

  
  
  
  

add a comment | 

2

$begingroup$

let $beta$ be half of the bisected angle, and $theta$ denote the angle AEB.

then the sine rule in triangle AEB gives:

$$
frac{|AE|}{sin beta} = frac{c}{sin theta}
$$

and the sine rule in triangle CEB gives:

$$
frac{|EC|}{sin beta} = frac{a}{sin (pi - theta)}
$$

since $sin theta = sin (pi-theta)$ and $|AE|+|EC| = b$, the result follows

share|cite|improve this answer

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

$endgroup$

add a comment | 

1

$begingroup$

The angle bisector theorem:
$$frac{CE}{AE}=frac ac$$
Then:
$$frac{bc}{a+c}=frac{b}{frac ac+1}=frac{b}{frac{CE}{AE}+1}=frac{AEcdot b}{CE+AE}=AE$$

share|cite|improve this answer

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Hint: We get $$frac{AE}{EC}=frac{c}{a}$$
And $$EC=b-AE$$ then we get $$frac{b-AE}{AE}=frac{c}{a}$$

share|cite|improve this answer

edited Feb 2 at 18:25

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you mean $frac ca$ ? But I still can't get to what I asked from here.
  $endgroup$
  – Pero
  Feb 2 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the angle bisector theorem
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
  $endgroup$
  – Pero
  Feb 2 at 18:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for that typo!
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:10
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: We get $$frac{AE}{EC}=frac{c}{a}$$
And $$EC=b-AE$$ then we get $$frac{b-AE}{AE}=frac{c}{a}$$

share|cite|improve this answer

edited Feb 2 at 18:25

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you mean $frac ca$ ? But I still can't get to what I asked from here.
  $endgroup$
  – Pero
  Feb 2 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the angle bisector theorem
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
  $endgroup$
  – Pero
  Feb 2 at 18:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for that typo!
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:10
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Hint: We get $$frac{AE}{EC}=frac{c}{a}$$
And $$EC=b-AE$$ then we get $$frac{b-AE}{AE}=frac{c}{a}$$

share|cite|improve this answer

edited Feb 2 at 18:25

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

$endgroup$

Hint: We get $$frac{AE}{EC}=frac{c}{a}$$
And $$EC=b-AE$$ then we get $$frac{b-AE}{AE}=frac{c}{a}$$

share|cite|improve this answer

edited Feb 2 at 18:25

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

share|cite|improve this answer

share|cite|improve this answer

edited Feb 2 at 18:25

edited Feb 2 at 18:25

edited Feb 2 at 18:25

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

answered Feb 2 at 18:01

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79k42867

79k42867

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you mean $frac ca$ ? But I still can't get to what I asked from here.
  $endgroup$
  – Pero
  Feb 2 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the angle bisector theorem
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
  $endgroup$
  – Pero
  Feb 2 at 18:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for that typo!
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:10
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you mean $frac ca$ ? But I still can't get to what I asked from here.
  $endgroup$
  – Pero
  Feb 2 at 18:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the angle bisector theorem
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
  $endgroup$
  – Pero
  Feb 2 at 18:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for that typo!
  $endgroup$
  – Dr. Sonnhard Graubner
  Feb 2 at 18:10
  

  
  
  
  

$begingroup$
you mean $frac ca$ ? But I still can't get to what I asked from here.
$endgroup$
– Pero
Feb 2 at 18:04

$begingroup$
you mean $frac ca$ ? But I still can't get to what I asked from here.
$endgroup$
– Pero
Feb 2 at 18:04

$begingroup$
It is the angle bisector theorem
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 18:08

$begingroup$
It is the angle bisector theorem
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 18:08

$begingroup$
Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
$endgroup$
– Pero
Feb 2 at 18:10

$begingroup$
Yes, I know it. I tried to get to what I asked with the angle bisector theorem, but I couldn't. What am I missing?
$endgroup$
– Pero
Feb 2 at 18:10

$begingroup$
Sorry for that typo!
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 18:10

$begingroup$
Sorry for that typo!
$endgroup$
– Dr. Sonnhard Graubner
Feb 2 at 18:10

add a comment | 

2

$begingroup$

let $beta$ be half of the bisected angle, and $theta$ denote the angle AEB.

then the sine rule in triangle AEB gives:

$$
frac{|AE|}{sin beta} = frac{c}{sin theta}
$$

and the sine rule in triangle CEB gives:

$$
frac{|EC|}{sin beta} = frac{a}{sin (pi - theta)}
$$

since $sin theta = sin (pi-theta)$ and $|AE|+|EC| = b$, the result follows

share|cite|improve this answer

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

$endgroup$

add a comment | 

2

$begingroup$

let $beta$ be half of the bisected angle, and $theta$ denote the angle AEB.

then the sine rule in triangle AEB gives:

$$
frac{|AE|}{sin beta} = frac{c}{sin theta}
$$

and the sine rule in triangle CEB gives:

$$
frac{|EC|}{sin beta} = frac{a}{sin (pi - theta)}
$$

since $sin theta = sin (pi-theta)$ and $|AE|+|EC| = b$, the result follows

share|cite|improve this answer

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

$endgroup$

add a comment | 

2

2

2

$begingroup$

let $beta$ be half of the bisected angle, and $theta$ denote the angle AEB.

then the sine rule in triangle AEB gives:

$$
frac{|AE|}{sin beta} = frac{c}{sin theta}
$$

and the sine rule in triangle CEB gives:

$$
frac{|EC|}{sin beta} = frac{a}{sin (pi - theta)}
$$

since $sin theta = sin (pi-theta)$ and $|AE|+|EC| = b$, the result follows

share|cite|improve this answer

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

$endgroup$

let $beta$ be half of the bisected angle, and $theta$ denote the angle AEB.

then the sine rule in triangle AEB gives:

$$
frac{|AE|}{sin beta} = frac{c}{sin theta}
$$

and the sine rule in triangle CEB gives:

$$
frac{|EC|}{sin beta} = frac{a}{sin (pi - theta)}
$$

since $sin theta = sin (pi-theta)$ and $|AE|+|EC| = b$, the result follows

share|cite|improve this answer

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

share|cite|improve this answer

share|cite|improve this answer

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

answered Feb 2 at 18:25

David HoldenDavid Holden

14.9k21225

14.9k21225

add a comment | 

add a comment | 

1

$begingroup$

The angle bisector theorem:
$$frac{CE}{AE}=frac ac$$
Then:
$$frac{bc}{a+c}=frac{b}{frac ac+1}=frac{b}{frac{CE}{AE}+1}=frac{AEcdot b}{CE+AE}=AE$$

share|cite|improve this answer

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

$endgroup$

add a comment | 

1

$begingroup$

The angle bisector theorem:
$$frac{CE}{AE}=frac ac$$
Then:
$$frac{bc}{a+c}=frac{b}{frac ac+1}=frac{b}{frac{CE}{AE}+1}=frac{AEcdot b}{CE+AE}=AE$$

share|cite|improve this answer

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

$endgroup$

add a comment | 

1

1

1

$begingroup$

The angle bisector theorem:
$$frac{CE}{AE}=frac ac$$
Then:
$$frac{bc}{a+c}=frac{b}{frac ac+1}=frac{b}{frac{CE}{AE}+1}=frac{AEcdot b}{CE+AE}=AE$$

share|cite|improve this answer

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

$endgroup$

The angle bisector theorem:
$$frac{CE}{AE}=frac ac$$
Then:
$$frac{bc}{a+c}=frac{b}{frac ac+1}=frac{b}{frac{CE}{AE}+1}=frac{AEcdot b}{CE+AE}=AE$$

share|cite|improve this answer

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

share|cite|improve this answer

share|cite|improve this answer

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

answered Feb 2 at 18:27

farruhotafarruhota

22.1k2942

22.1k2942

add a comment | 

add a comment | 

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But avoid …

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• 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.

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