In a triangle $angle A = 2angle B$ iff $a^2 = b(b+c)$

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Prove that in a triangle $ABC$, $angle A = angle 2B$, if and only if:

$$a^2 = b(b+c)$$

where $a, b, c$ are the sides opposite to $A, B, C$ respectively.

I attacked the problem using the Law of Sines, and tried to prove that if $angle A$ was indeed equal to $2angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof.

From the Law of Sines,

$$a = 2Rsin A = 2Rsin (2B) = 4Rsin Bcos B$$

$$b = 2Rsin B$$

$$c = 2Rsin C = 2Rsin(180 - 3B) = 2Rsin(3B) = 2R(sin Bcos(2B) + sin(2B)cos B)$$

$$=2R(sin B(1 - 2sin^2 B) +2sin Bcos^2 B) = 2R(sin B -2sin^3 B + 2sin B(1 - sin^2B))$$

$$=boxed2R(3sin B - 4sin^3 B)$$

Now,

$$implies b(b+c) = 2Rsin B[2Rsin B + 2R(3sin B - 4sin^3 B)]$$

$$=4R^2sin^2 B(1 + 3 - 4sin^2 B)$$

$$=16R^2sin^2 Bcos^2 B = a^2$$

Now, to prove the converse:

$$c = 2Rsin C = 2Rsin (180 - (A + B)) = 2Rsin(A+B) = 2Rsin Acos B + 2Rsin Bcos A$$

$$a^2 = b(b+c)$$

$$implies 4R^2sin^2 A = 2Rsin B(2Rsin B + 2Rsin Acos B + 2Rsin Bcos) $$

$$ = 4R^2sin B(sin B + sin Acos B + sin Bcos A)$$

I have no idea how to proceed from here. I tried replacing $sin A$ with $sqrt1 - cos^2 B$, but that doesn't yield any useful results.

asked Nov 9 '13 at 7:00

Gerard

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up vote
5
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favorite

Prove that in a triangle $ABC$, $angle A = angle 2B$, if and only if:

$$a^2 = b(b+c)$$

where $a, b, c$ are the sides opposite to $A, B, C$ respectively.

From the Law of Sines,

$$a = 2Rsin A = 2Rsin (2B) = 4Rsin Bcos B$$

$$b = 2Rsin B$$

$$c = 2Rsin C = 2Rsin(180 - 3B) = 2Rsin(3B) = 2R(sin Bcos(2B) + sin(2B)cos B)$$

$$=2R(sin B(1 - 2sin^2 B) +2sin Bcos^2 B) = 2R(sin B -2sin^3 B + 2sin B(1 - sin^2B))$$

$$=boxed2R(3sin B - 4sin^3 B)$$

Now,

$$implies b(b+c) = 2Rsin B[2Rsin B + 2R(3sin B - 4sin^3 B)]$$

$$=4R^2sin^2 B(1 + 3 - 4sin^2 B)$$

$$=16R^2sin^2 Bcos^2 B = a^2$$

Now, to prove the converse:

$$c = 2Rsin C = 2Rsin (180 - (A + B)) = 2Rsin(A+B) = 2Rsin Acos B + 2Rsin Bcos A$$

$$a^2 = b(b+c)$$

$$implies 4R^2sin^2 A = 2Rsin B(2Rsin B + 2Rsin Acos B + 2Rsin Bcos) $$

$$ = 4R^2sin B(sin B + sin Acos B + sin Bcos A)$$

I have no idea how to proceed from here. I tried replacing $sin A$ with $sqrt1 - cos^2 B$, but that doesn't yield any useful results.

asked Nov 9 '13 at 7:00

Gerard

1,97021333

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up vote
5
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favorite

Prove that in a triangle $ABC$, $angle A = angle 2B$, if and only if:

$$a^2 = b(b+c)$$

where $a, b, c$ are the sides opposite to $A, B, C$ respectively.

From the Law of Sines,

$$a = 2Rsin A = 2Rsin (2B) = 4Rsin Bcos B$$

$$b = 2Rsin B$$

$$c = 2Rsin C = 2Rsin(180 - 3B) = 2Rsin(3B) = 2R(sin Bcos(2B) + sin(2B)cos B)$$

$$=2R(sin B(1 - 2sin^2 B) +2sin Bcos^2 B) = 2R(sin B -2sin^3 B + 2sin B(1 - sin^2B))$$

$$=boxed2R(3sin B - 4sin^3 B)$$

Now,

$$implies b(b+c) = 2Rsin B[2Rsin B + 2R(3sin B - 4sin^3 B)]$$

$$=4R^2sin^2 B(1 + 3 - 4sin^2 B)$$

$$=16R^2sin^2 Bcos^2 B = a^2$$

Now, to prove the converse:

$$c = 2Rsin C = 2Rsin (180 - (A + B)) = 2Rsin(A+B) = 2Rsin Acos B + 2Rsin Bcos A$$

$$a^2 = b(b+c)$$

$$implies 4R^2sin^2 A = 2Rsin B(2Rsin B + 2Rsin Acos B + 2Rsin Bcos) $$

$$ = 4R^2sin B(sin B + sin Acos B + sin Bcos A)$$

I have no idea how to proceed from here. I tried replacing $sin A$ with $sqrt1 - cos^2 B$, but that doesn't yield any useful results.

asked Nov 9 '13 at 7:00

Gerard

1,97021333

Prove that in a triangle $ABC$, $angle A = angle 2B$, if and only if:

$$a^2 = b(b+c)$$

where $a, b, c$ are the sides opposite to $A, B, C$ respectively.

From the Law of Sines,

$$a = 2Rsin A = 2Rsin (2B) = 4Rsin Bcos B$$

$$b = 2Rsin B$$

$$c = 2Rsin C = 2Rsin(180 - 3B) = 2Rsin(3B) = 2R(sin Bcos(2B) + sin(2B)cos B)$$

$$=2R(sin B(1 - 2sin^2 B) +2sin Bcos^2 B) = 2R(sin B -2sin^3 B + 2sin B(1 - sin^2B))$$

$$=boxed2R(3sin B - 4sin^3 B)$$

Now,

$$implies b(b+c) = 2Rsin B[2Rsin B + 2R(3sin B - 4sin^3 B)]$$

$$=4R^2sin^2 B(1 + 3 - 4sin^2 B)$$

$$=16R^2sin^2 Bcos^2 B = a^2$$

Now, to prove the converse:

$$c = 2Rsin C = 2Rsin (180 - (A + B)) = 2Rsin(A+B) = 2Rsin Acos B + 2Rsin Bcos A$$

$$a^2 = b(b+c)$$

$$implies 4R^2sin^2 A = 2Rsin B(2Rsin B + 2Rsin Acos B + 2Rsin Bcos) $$

$$ = 4R^2sin B(sin B + sin Acos B + sin Bcos A)$$

I have no idea how to proceed from here. I tried replacing $sin A$ with $sqrt1 - cos^2 B$, but that doesn't yield any useful results.

asked Nov 9 '13 at 7:00

Gerard

1,97021333

asked Nov 9 '13 at 7:00

Gerard

1,97021333

asked Nov 9 '13 at 7:00

Gerard

1,97021333

asked Nov 9 '13 at 7:00

Gerard

1,97021333

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3 Answers
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accepted

$$a^2-b^2=bcimplies sin^2A-sin^2B=sin Bsin Ctext as Rne0$$

Now, $displaystylesin^2A-sin^2B=sin(A+B)sin(A-B)=sin(pi-C)sin(A-B)=sin Csin(A-B) (1)$

$$implies sin Bsin C=sin Csin(A-B)$$

$$implies sin B=sin(A-B)text as sin Cne0$$

$$implies B=npi+(-1)^n(A-B)text where ntext is any integer $$

If $n$ is even, $n=2m$(say) $implies B=2mpi+A-Biff A=2B-2mpi=2B$ as $0<A,B<pi$

If $n$ is odd, $n=2m+1$(say) $implies B=(2m+1)pi-(A-B)iff A=(2m+1)pi$ which is impossible as $0<A<pi$

Conversely, if $A=2B$

$displaystyleimplies a^2-b^2=4R^2(sin^2A-sin^2B)=4R^2sin Csin(A-B)$ (using $(1)$)

$displaystyle=4R^2sin Csin(2B-B)=2Rsin Bcdot 2Rsin C=cdots$

edited Nov 9 '13 at 7:14

answered Nov 9 '13 at 7:09

lab bhattacharjee

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enter image description here

1) $angle A=2angle B$

angle bisector of $angle A$ cut $BC$ at $D$ $Longrightarrow$ $dfracCDDB=dfracbc$

thus, $CD=dfracbb+ca$

also, $triangle CADsimtriangle CAB$ $Longrightarrow$ $CA^2=CDcdot CB$

hence, $b^2=dfracbb+cacdot a$ $Longrightarrow$ $a^2=b(b+c)$

2) $a^2=b(b+c)$

angle bisector of $angle A$ cut $BC$ at $D$ $Longrightarrow$ $dfracCDDB=dfracbc$ $Longrightarrow$ $CD=dfracbb+ca$

$a^2=b(b+c)$ $Longrightarrow$ $dfracab=dfracb+ca=dfracbdfracbb+ca$ $Longrightarrow$ $dfracBCCA=dfracCACD$ and $angle DCA=angle ACB$

thus, $triangle ACDsimtriangle ACB$ $Longrightarrow$ $angle CAD=angle CBA$ $Longrightarrow$ $angle A=2angle B$

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

Very elegant solution. How did you produce the drawing?
â€“Â Gerard
Nov 9 '13 at 9:12

This answer can be streamlined a bit. Regardless of the bisector status of $overlineAD$ (but still assuming $overlineAD$ in $angle BAC$'s interior), we have $angle BAC cong angle ADC$, so that $triangle BAC sim angle ADC$ and $$fracb = fracba$$ Then $$beginalign overlineAD text bisects angle A quad &Leftrightarrow qquad fracb = fraca-c \[4pt] &Leftrightarrow qquad fracb^2/ab = fraca-b^2/ac \[4pt] &Leftrightarrow qquad a^2 = bleft( b + c right ) endalign$$
â€“Â Blue
Nov 10 '13 at 2:18

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Double angle formula says
$$
beginalign
sin(A)
&=2sin(B)cos(B)\
&impliesfracsin(A)sin(B)=2cos(B)tag1
endalign
$$
The formula for the sine of a sum yields
$$
beginalign
sin(C)
&=sin(A+B)\
&=2sin(B)cos(B)cos(B)+(2cos^2(B)-1)sin(B)\
&=sin(B)(4cos^2(B)-1)\
&impliesfracsin(C)sin(B)=4cos^2(B)-1tag2
endalign
$$
Thus, the Law of Sines says
$$
left(frac abright)^2-1=frac cbimplies a^2-b^2=bctag3
$$

answered Mar 12 '14 at 21:43

robjohnâ™¦

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3 Answers
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3 Answers
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active

oldest

votes

up vote
2
down vote

accepted

$$a^2-b^2=bcimplies sin^2A-sin^2B=sin Bsin Ctext as Rne0$$

Now, $displaystylesin^2A-sin^2B=sin(A+B)sin(A-B)=sin(pi-C)sin(A-B)=sin Csin(A-B) (1)$

$$implies sin Bsin C=sin Csin(A-B)$$

$$implies sin B=sin(A-B)text as sin Cne0$$

$$implies B=npi+(-1)^n(A-B)text where ntext is any integer $$

If $n$ is even, $n=2m$(say) $implies B=2mpi+A-Biff A=2B-2mpi=2B$ as $0<A,B<pi$

If $n$ is odd, $n=2m+1$(say) $implies B=(2m+1)pi-(A-B)iff A=(2m+1)pi$ which is impossible as $0<A<pi$

Conversely, if $A=2B$

$displaystyleimplies a^2-b^2=4R^2(sin^2A-sin^2B)=4R^2sin Csin(A-B)$ (using $(1)$)

$displaystyle=4R^2sin Csin(2B-B)=2Rsin Bcdot 2Rsin C=cdots$

edited Nov 9 '13 at 7:14

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

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up vote
2
down vote

accepted

$$a^2-b^2=bcimplies sin^2A-sin^2B=sin Bsin Ctext as Rne0$$

Now, $displaystylesin^2A-sin^2B=sin(A+B)sin(A-B)=sin(pi-C)sin(A-B)=sin Csin(A-B) (1)$

$$implies sin Bsin C=sin Csin(A-B)$$

$$implies sin B=sin(A-B)text as sin Cne0$$

$$implies B=npi+(-1)^n(A-B)text where ntext is any integer $$

If $n$ is even, $n=2m$(say) $implies B=2mpi+A-Biff A=2B-2mpi=2B$ as $0<A,B<pi$

If $n$ is odd, $n=2m+1$(say) $implies B=(2m+1)pi-(A-B)iff A=(2m+1)pi$ which is impossible as $0<A<pi$

Conversely, if $A=2B$

$displaystyleimplies a^2-b^2=4R^2(sin^2A-sin^2B)=4R^2sin Csin(A-B)$ (using $(1)$)

$displaystyle=4R^2sin Csin(2B-B)=2Rsin Bcdot 2Rsin C=cdots$

edited Nov 9 '13 at 7:14

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

add a commentÂ |Â

up vote
2
down vote

accepted

$$a^2-b^2=bcimplies sin^2A-sin^2B=sin Bsin Ctext as Rne0$$

Now, $displaystylesin^2A-sin^2B=sin(A+B)sin(A-B)=sin(pi-C)sin(A-B)=sin Csin(A-B) (1)$

$$implies sin Bsin C=sin Csin(A-B)$$

$$implies sin B=sin(A-B)text as sin Cne0$$

$$implies B=npi+(-1)^n(A-B)text where ntext is any integer $$

If $n$ is even, $n=2m$(say) $implies B=2mpi+A-Biff A=2B-2mpi=2B$ as $0<A,B<pi$

If $n$ is odd, $n=2m+1$(say) $implies B=(2m+1)pi-(A-B)iff A=(2m+1)pi$ which is impossible as $0<A<pi$

Conversely, if $A=2B$

$displaystyleimplies a^2-b^2=4R^2(sin^2A-sin^2B)=4R^2sin Csin(A-B)$ (using $(1)$)

$displaystyle=4R^2sin Csin(2B-B)=2Rsin Bcdot 2Rsin C=cdots$

edited Nov 9 '13 at 7:14

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

$$a^2-b^2=bcimplies sin^2A-sin^2B=sin Bsin Ctext as Rne0$$

Now, $displaystylesin^2A-sin^2B=sin(A+B)sin(A-B)=sin(pi-C)sin(A-B)=sin Csin(A-B) (1)$

$$implies sin Bsin C=sin Csin(A-B)$$

$$implies sin B=sin(A-B)text as sin Cne0$$

$$implies B=npi+(-1)^n(A-B)text where ntext is any integer $$

If $n$ is even, $n=2m$(say) $implies B=2mpi+A-Biff A=2B-2mpi=2B$ as $0<A,B<pi$

If $n$ is odd, $n=2m+1$(say) $implies B=(2m+1)pi-(A-B)iff A=(2m+1)pi$ which is impossible as $0<A<pi$

Conversely, if $A=2B$

$displaystyleimplies a^2-b^2=4R^2(sin^2A-sin^2B)=4R^2sin Csin(A-B)$ (using $(1)$)

$displaystyle=4R^2sin Csin(2B-B)=2Rsin Bcdot 2Rsin C=cdots$

edited Nov 9 '13 at 7:14

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

edited Nov 9 '13 at 7:14

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

answered Nov 9 '13 at 7:09

lab bhattacharjee

215k14152264

add a commentÂ |Â

up vote
8
down vote

enter image description here

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

Very elegant solution. How did you produce the drawing?
â€“Â Gerard
Nov 9 '13 at 9:12

This answer can be streamlined a bit. Regardless of the bisector status of $overlineAD$ (but still assuming $overlineAD$ in $angle BAC$'s interior), we have $angle BAC cong angle ADC$, so that $triangle BAC sim angle ADC$ and $$fracb = fracba$$ Then $$beginalign overlineAD text bisects angle A quad &Leftrightarrow qquad fracb = fraca-c \[4pt] &Leftrightarrow qquad fracb^2/ab = fraca-b^2/ac \[4pt] &Leftrightarrow qquad a^2 = bleft( b + c right ) endalign$$
â€“Â Blue
Nov 10 '13 at 2:18

add a commentÂ |Â

up vote
8
down vote

enter image description here

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

Very elegant solution. How did you produce the drawing?
â€“Â Gerard
Nov 9 '13 at 9:12

This answer can be streamlined a bit. Regardless of the bisector status of $overlineAD$ (but still assuming $overlineAD$ in $angle BAC$'s interior), we have $angle BAC cong angle ADC$, so that $triangle BAC sim angle ADC$ and $$fracb = fracba$$ Then $$beginalign overlineAD text bisects angle A quad &Leftrightarrow qquad fracb = fraca-c \[4pt] &Leftrightarrow qquad fracb^2/ab = fraca-b^2/ac \[4pt] &Leftrightarrow qquad a^2 = bleft( b + c right ) endalign$$
â€“Â Blue
Nov 10 '13 at 2:18

add a commentÂ |Â

up vote
8
down vote

enter image description here

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

enter image description here

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

answered Nov 9 '13 at 8:43

chloe_shi

1,947712

Very elegant solution. How did you produce the drawing?
â€“Â Gerard
Nov 9 '13 at 9:12

This answer can be streamlined a bit. Regardless of the bisector status of $overlineAD$ (but still assuming $overlineAD$ in $angle BAC$'s interior), we have $angle BAC cong angle ADC$, so that $triangle BAC sim angle ADC$ and $$fracb = fracba$$ Then $$beginalign overlineAD text bisects angle A quad &Leftrightarrow qquad fracb = fraca-c \[4pt] &Leftrightarrow qquad fracb^2/ab = fraca-b^2/ac \[4pt] &Leftrightarrow qquad a^2 = bleft( b + c right ) endalign$$
â€“Â Blue
Nov 10 '13 at 2:18

add a commentÂ |Â

Very elegant solution. How did you produce the drawing?
â€“Â Gerard
Nov 9 '13 at 9:12

This answer can be streamlined a bit. Regardless of the bisector status of $overlineAD$ (but still assuming $overlineAD$ in $angle BAC$'s interior), we have $angle BAC cong angle ADC$, so that $triangle BAC sim angle ADC$ and $$fracb = fracba$$ Then $$beginalign overlineAD text bisects angle A quad &Leftrightarrow qquad fracb = fraca-c \[4pt] &Leftrightarrow qquad fracb^2/ab = fraca-b^2/ac \[4pt] &Leftrightarrow qquad a^2 = bleft( b + c right ) endalign$$
â€“Â Blue
Nov 10 '13 at 2:18

Very elegant solution. How did you produce the drawing?
â€“Â Gerard
Nov 9 '13 at 9:12

This answer can be streamlined a bit. Regardless of the bisector status of $overlineAD$ (but still assuming $overlineAD$ in $angle BAC$'s interior), we have $angle BAC cong angle ADC$, so that $triangle BAC sim angle ADC$ and $$fracb = fracba$$ Then $$beginalign overlineAD text bisects angle A quad &Leftrightarrow qquad fracb = fraca-c \[4pt] &Leftrightarrow qquad fracb^2/ab = fraca-b^2/ac \[4pt] &Leftrightarrow qquad a^2 = bleft( b + c right ) endalign$$
â€“Â Blue
Nov 10 '13 at 2:18

add a commentÂ |Â

up vote
2
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answered Mar 12 '14 at 21:43

robjohnâ™¦

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answered Mar 12 '14 at 21:43

robjohnâ™¦

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answered Mar 12 '14 at 21:43

robjohnâ™¦

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answered Mar 12 '14 at 21:43

robjohnâ™¦

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answered Mar 12 '14 at 21:43

robjohnâ™¦

258k26297612

answered Mar 12 '14 at 21:43

robjohnâ™¦

258k26297612

answered Mar 12 '14 at 21:43

robjohnâ™¦

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