Simpler Methods Geometry
$begingroup$
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac{4x^2}{121} - frac{4x^2cos A}{11}$
$BD = frac{2BC}{11}$
$frac{sin2A}{CD} = frac{sin CAD}{11}$
$frac{sin A}{CD} = frac{sin CDB}{x}$
Solving these systems would give us 13.
geometry
edited Jul 18 '18 at 18:48
GuySa
364313
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
$endgroup$
add a comment |
$begingroup$
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac{4x^2}{121} - frac{4x^2cos A}{11}$
$BD = frac{2BC}{11}$
$frac{sin2A}{CD} = frac{sin CAD}{11}$
$frac{sin A}{CD} = frac{sin CDB}{x}$
Solving these systems would give us 13.
geometry
edited Jul 18 '18 at 18:48
GuySa
364313
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
$endgroup$
$begingroup$
Could you attach the figure for this question? It's not clear what is $D$.
$endgroup$
– Jaroslaw Matlak
Jul 18 '18 at 15:13
$begingroup$
Try Stewart's Theorem.
$endgroup$
– MalayTheDynamo
Jul 18 '18 at 15:19
$begingroup$
Im trying but what would you do after getting CD in terms of BC?
$endgroup$
– SuperMage1
Jul 18 '18 at 15:27
add a comment |
$begingroup$
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac{4x^2}{121} - frac{4x^2cos A}{11}$
$BD = frac{2BC}{11}$
$frac{sin2A}{CD} = frac{sin CAD}{11}$
$frac{sin A}{CD} = frac{sin CDB}{x}$
Solving these systems would give us 13.
geometry
edited Jul 18 '18 at 18:48
GuySa
364313
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
$endgroup$
In $△ABC$, $∠A = 2∠B$ , $CD$ bisects $∠ACB$ , $AC =11cm$ and $AD = 2cm$. Find the length, in cm, of $BC$. Attached is the figure of the question
I already solved it using the Angle Bisector Theorem and sine and cosine law, but is pretty tedious to do. I am asking for solutions that would be a lot quicker.
Solution:
My system of equations:
$ 11^2 +2^2 -44cos2A = x^2 + frac{4x^2}{121} - frac{4x^2cos A}{11}$
$BD = frac{2BC}{11}$
$frac{sin2A}{CD} = frac{sin CAD}{11}$
$frac{sin A}{CD} = frac{sin CDB}{x}$
Solving these systems would give us 13.
geometry
geometry
edited Jul 18 '18 at 18:48
GuySa
364313
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
edited Jul 18 '18 at 18:48
GuySa
364313
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
edited Jul 18 '18 at 18:48
GuySa
364313
edited Jul 18 '18 at 18:48
GuySa
364313
edited Jul 18 '18 at 18:48
GuySa
364313
364313
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
asked Jul 18 '18 at 15:00
SuperMage1SuperMage1
897211
897211
$begingroup$
Could you attach the figure for this question? It's not clear what is $D$.
$endgroup$
– Jaroslaw Matlak
Jul 18 '18 at 15:13
$begingroup$
Try Stewart's Theorem.
$endgroup$
– MalayTheDynamo
Jul 18 '18 at 15:19
$begingroup$
Im trying but what would you do after getting CD in terms of BC?
$endgroup$
– SuperMage1
Jul 18 '18 at 15:27
add a comment |
$begingroup$
Could you attach the figure for this question? It's not clear what is $D$.
$endgroup$
– Jaroslaw Matlak
Jul 18 '18 at 15:13
$begingroup$
Try Stewart's Theorem.
$endgroup$
– MalayTheDynamo
Jul 18 '18 at 15:19
$begingroup$
Im trying but what would you do after getting CD in terms of BC?
$endgroup$
– SuperMage1
Jul 18 '18 at 15:27
$begingroup$
Could you attach the figure for this question? It's not clear what is $D$.
$endgroup$
– Jaroslaw Matlak
Jul 18 '18 at 15:13
$begingroup$
Could you attach the figure for this question? It's not clear what is $D$.
$endgroup$
– Jaroslaw Matlak
Jul 18 '18 at 15:13
$begingroup$
Try Stewart's Theorem.
$endgroup$
– MalayTheDynamo
Jul 18 '18 at 15:19
$begingroup$
Try Stewart's Theorem.
$endgroup$
– MalayTheDynamo
Jul 18 '18 at 15:19
$begingroup$
Im trying but what would you do after getting CD in terms of BC?
$endgroup$
– SuperMage1
Jul 18 '18 at 15:27
$begingroup$
Im trying but what would you do after getting CD in terms of BC?
$endgroup$
– SuperMage1
Jul 18 '18 at 15:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$.
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is the $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac{180-2x}{2} = frac{3y}{2}$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac{3y}{2} = {yover 2}$$ which implies $angle OMD$ also equals $y/2$ (CPCT).
Also , $DM = AD = 2$ (equal base angles).
We have,
$$angle MDB = 2*{yover 2} = y$$
Therefore , $MD = MB = 2$ (equal base angles).
Therefore, $$BC = MC + MB = 11+2 = boxed{13} $$
answered Jul 21 '18 at 12:10
SinπSinπ
64511
$endgroup$
add a comment |
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1 Answer
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$begingroup$
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$.
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is the $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac{180-2x}{2} = frac{3y}{2}$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac{3y}{2} = {yover 2}$$ which implies $angle OMD$ also equals $y/2$ (CPCT).
Also , $DM = AD = 2$ (equal base angles).
We have,
$$angle MDB = 2*{yover 2} = y$$
Therefore , $MD = MB = 2$ (equal base angles).
Therefore, $$BC = MC + MB = 11+2 = boxed{13} $$
answered Jul 21 '18 at 12:10
SinπSinπ
64511
$endgroup$
add a comment |
$begingroup$
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$.
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is the $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac{180-2x}{2} = frac{3y}{2}$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac{3y}{2} = {yover 2}$$ which implies $angle OMD$ also equals $y/2$ (CPCT).
Also , $DM = AD = 2$ (equal base angles).
We have,
$$angle MDB = 2*{yover 2} = y$$
Therefore , $MD = MB = 2$ (equal base angles).
Therefore, $$BC = MC + MB = 11+2 = boxed{13} $$
answered Jul 21 '18 at 12:10
SinπSinπ
64511
$endgroup$
add a comment |
$begingroup$
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$.
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is the $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac{180-2x}{2} = frac{3y}{2}$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac{3y}{2} = {yover 2}$$ which implies $angle OMD$ also equals $y/2$ (CPCT).
Also , $DM = AD = 2$ (equal base angles).
We have,
$$angle MDB = 2*{yover 2} = y$$
Therefore , $MD = MB = 2$ (equal base angles).
Therefore, $$BC = MC + MB = 11+2 = boxed{13} $$
answered Jul 21 '18 at 12:10
SinπSinπ
64511
$endgroup$
This is another way to do it.
Locate a point $M$ on $BC$ , such that $MC$ equals $11$. Join $AM$ , $MD$. Let the intersection of $AM$ and $CD$ be $O$.
For convenience , let $$angle ACD = angle DCB = x \∠CAD = 2y\ ∠CBD = y$$ where $DC$ is the $angle$ bisector.
$triangle ACM$ is isosceles. This implies the base angles are equal and are $$frac{180-2x}{2} = frac{3y}{2}$$
(Since $180-2x = 3y$).
Also , $triangle$s $AOC$ and $MOC$ are congruent. This further implies $triangle AOD$ and $triangle MOD$ are congruent (SAS).
$$angle OAD = 2y- frac{3y}{2} = {yover 2}$$ which implies $angle OMD$ also equals $y/2$ (CPCT).
Also , $DM = AD = 2$ (equal base angles).
We have,
$$angle MDB = 2*{yover 2} = y$$
Therefore , $MD = MB = 2$ (equal base angles).
Therefore, $$BC = MC + MB = 11+2 = boxed{13} $$
answered Jul 21 '18 at 12:10
SinπSinπ
64511
edited Dec 30 '18 at 9:29
answered Jul 21 '18 at 12:10
SinπSinπ
64511
answered Jul 21 '18 at 12:10
SinπSinπ
64511
answered Jul 21 '18 at 12:10
SinπSinπ
64511
64511
add a comment |
add a comment |
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$begingroup$
Could you attach the figure for this question? It's not clear what is $D$.
$endgroup$
– Jaroslaw Matlak
Jul 18 '18 at 15:13
$begingroup$
Try Stewart's Theorem.
$endgroup$
– MalayTheDynamo
Jul 18 '18 at 15:19
$begingroup$
Im trying but what would you do after getting CD in terms of BC?
$endgroup$
– SuperMage1
Jul 18 '18 at 15:27