In $triangle ABC$ with $D$ on $overline{AC}$, if $angle CBD=2angle ABD$ and the circumcenter lies on...
$begingroup$
- Let $alpha$ be $measuredangle ABD$
- Let $beta$ be $measuredangle DBC$
- Let D be a point on AC such that BD passes through the origin point O
Prove that $frac{AD}{DC}$ cannot be equal to $frac{1}{2}$ when $alpha = frac{1}{2}beta$
Here's what I have:
$$measuredangle AOD = 2alpha$$
$$measuredangle DOC = 2beta$$
From this information I completed all the angles and reached the following (correct) equation by using the law of sines on triangles AOD and DOC, which share the equal length OC = AO:
$$frac{AD}{DC} = frac{sin 2beta}{sin 2alpha}$$
Given that $alpha = frac{1}{2}beta$:
$$frac{AD}{DC} = frac{sin 2beta}{sin beta} = frac{2sin beta cos beta}{sin beta} = 2cosbeta$$
$$downarrow$$
$$2cosbeta = 1/2$$
$$downarrow$$
$$beta = 75.52^circ$$
If $beta = 75.52^circ$, then $2alpha + 2beta > 180^circ$, and the triangle will now look like this:
In this situation, BD cannot pass through O, which breaks the definition of the problem.
This is the best proof I can come up with. I tried first to prove it numerically by summing up angles to 180, but that did not work as all the statements were true. I feel that my proof is borderline illegal and that I do not address all cases, so I am asking if anyone could figure out a more elegant, preferably algebraic alternative.
geometry trigonometry euclidean-geometry circles
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
$endgroup$
add a comment |
$begingroup$
- Let $alpha$ be $measuredangle ABD$
- Let $beta$ be $measuredangle DBC$
- Let D be a point on AC such that BD passes through the origin point O
Prove that $frac{AD}{DC}$ cannot be equal to $frac{1}{2}$ when $alpha = frac{1}{2}beta$
Here's what I have:
$$measuredangle AOD = 2alpha$$
$$measuredangle DOC = 2beta$$
From this information I completed all the angles and reached the following (correct) equation by using the law of sines on triangles AOD and DOC, which share the equal length OC = AO:
$$frac{AD}{DC} = frac{sin 2beta}{sin 2alpha}$$
Given that $alpha = frac{1}{2}beta$:
$$frac{AD}{DC} = frac{sin 2beta}{sin beta} = frac{2sin beta cos beta}{sin beta} = 2cosbeta$$
$$downarrow$$
$$2cosbeta = 1/2$$
$$downarrow$$
$$beta = 75.52^circ$$
If $beta = 75.52^circ$, then $2alpha + 2beta > 180^circ$, and the triangle will now look like this:
In this situation, BD cannot pass through O, which breaks the definition of the problem.
This is the best proof I can come up with. I tried first to prove it numerically by summing up angles to 180, but that did not work as all the statements were true. I feel that my proof is borderline illegal and that I do not address all cases, so I am asking if anyone could figure out a more elegant, preferably algebraic alternative.
geometry trigonometry euclidean-geometry circles
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
$endgroup$
$begingroup$
@Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1.
$endgroup$
– daedsidog
Jan 4 at 15:15
1
$begingroup$
Whoops. .......
$endgroup$
– Blue
Jan 4 at 15:17
add a comment |
$begingroup$
- Let $alpha$ be $measuredangle ABD$
- Let $beta$ be $measuredangle DBC$
- Let D be a point on AC such that BD passes through the origin point O
Prove that $frac{AD}{DC}$ cannot be equal to $frac{1}{2}$ when $alpha = frac{1}{2}beta$
Here's what I have:
$$measuredangle AOD = 2alpha$$
$$measuredangle DOC = 2beta$$
From this information I completed all the angles and reached the following (correct) equation by using the law of sines on triangles AOD and DOC, which share the equal length OC = AO:
$$frac{AD}{DC} = frac{sin 2beta}{sin 2alpha}$$
Given that $alpha = frac{1}{2}beta$:
$$frac{AD}{DC} = frac{sin 2beta}{sin beta} = frac{2sin beta cos beta}{sin beta} = 2cosbeta$$
$$downarrow$$
$$2cosbeta = 1/2$$
$$downarrow$$
$$beta = 75.52^circ$$
If $beta = 75.52^circ$, then $2alpha + 2beta > 180^circ$, and the triangle will now look like this:
In this situation, BD cannot pass through O, which breaks the definition of the problem.
This is the best proof I can come up with. I tried first to prove it numerically by summing up angles to 180, but that did not work as all the statements were true. I feel that my proof is borderline illegal and that I do not address all cases, so I am asking if anyone could figure out a more elegant, preferably algebraic alternative.
geometry trigonometry euclidean-geometry circles
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
$endgroup$
- Let $alpha$ be $measuredangle ABD$
- Let $beta$ be $measuredangle DBC$
- Let D be a point on AC such that BD passes through the origin point O
Prove that $frac{AD}{DC}$ cannot be equal to $frac{1}{2}$ when $alpha = frac{1}{2}beta$
Here's what I have:
$$measuredangle AOD = 2alpha$$
$$measuredangle DOC = 2beta$$
From this information I completed all the angles and reached the following (correct) equation by using the law of sines on triangles AOD and DOC, which share the equal length OC = AO:
$$frac{AD}{DC} = frac{sin 2beta}{sin 2alpha}$$
Given that $alpha = frac{1}{2}beta$:
$$frac{AD}{DC} = frac{sin 2beta}{sin beta} = frac{2sin beta cos beta}{sin beta} = 2cosbeta$$
$$downarrow$$
$$2cosbeta = 1/2$$
$$downarrow$$
$$beta = 75.52^circ$$
If $beta = 75.52^circ$, then $2alpha + 2beta > 180^circ$, and the triangle will now look like this:
In this situation, BD cannot pass through O, which breaks the definition of the problem.
This is the best proof I can come up with. I tried first to prove it numerically by summing up angles to 180, but that did not work as all the statements were true. I feel that my proof is borderline illegal and that I do not address all cases, so I am asking if anyone could figure out a more elegant, preferably algebraic alternative.
geometry trigonometry euclidean-geometry circles
geometry trigonometry euclidean-geometry circles
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
edited Jan 4 at 15:26
Michael Rozenberg
109k1896201
109k1896201
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
asked Jan 4 at 14:37
daedsidogdaedsidog
31617
31617
$begingroup$
@Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1.
$endgroup$
– daedsidog
Jan 4 at 15:15
1
$begingroup$
Whoops. .......
$endgroup$
– Blue
Jan 4 at 15:17
add a comment |
$begingroup$
@Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1.
$endgroup$
– daedsidog
Jan 4 at 15:15
1
$begingroup$
Whoops. .......
$endgroup$
– Blue
Jan 4 at 15:17
$begingroup$
@Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1.
$endgroup$
– daedsidog
Jan 4 at 15:15
$begingroup$
@Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1.
$endgroup$
– daedsidog
Jan 4 at 15:15
1
1
$begingroup$
Whoops. .......
$endgroup$
– Blue
Jan 4 at 15:17
$begingroup$
Whoops. .......
$endgroup$
– Blue
Jan 4 at 15:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $frac{AD}{DC}=frac{1}{2}$ by the law of sines we obtain:
$$frac{1}{2}=frac{AD}{DC}=frac{frac{AD}{BD}}{frac{DC}{BD}}=frac{frac{sinalpha}{sinmeasuredangle A}}{frac{sin2alpha}{sinmeasuredangle C}}=frac{frac{sinalpha}{sin(90^{circ}-2a)}}{frac{sin2alpha}{sin(90^{circ}-alpha)}}=frac{sinalphacosalpha}{cos2alphasin2alpha}=frac{1}{2cos2alpha}.$$
Id est, $cos2alpha=1,$ which is impossible.
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
add a comment |
Your Answer
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$begingroup$
If $frac{AD}{DC}=frac{1}{2}$ by the law of sines we obtain:
$$frac{1}{2}=frac{AD}{DC}=frac{frac{AD}{BD}}{frac{DC}{BD}}=frac{frac{sinalpha}{sinmeasuredangle A}}{frac{sin2alpha}{sinmeasuredangle C}}=frac{frac{sinalpha}{sin(90^{circ}-2a)}}{frac{sin2alpha}{sin(90^{circ}-alpha)}}=frac{sinalphacosalpha}{cos2alphasin2alpha}=frac{1}{2cos2alpha}.$$
Id est, $cos2alpha=1,$ which is impossible.
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
add a comment |
$begingroup$
If $frac{AD}{DC}=frac{1}{2}$ by the law of sines we obtain:
$$frac{1}{2}=frac{AD}{DC}=frac{frac{AD}{BD}}{frac{DC}{BD}}=frac{frac{sinalpha}{sinmeasuredangle A}}{frac{sin2alpha}{sinmeasuredangle C}}=frac{frac{sinalpha}{sin(90^{circ}-2a)}}{frac{sin2alpha}{sin(90^{circ}-alpha)}}=frac{sinalphacosalpha}{cos2alphasin2alpha}=frac{1}{2cos2alpha}.$$
Id est, $cos2alpha=1,$ which is impossible.
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
add a comment |
$begingroup$
If $frac{AD}{DC}=frac{1}{2}$ by the law of sines we obtain:
$$frac{1}{2}=frac{AD}{DC}=frac{frac{AD}{BD}}{frac{DC}{BD}}=frac{frac{sinalpha}{sinmeasuredangle A}}{frac{sin2alpha}{sinmeasuredangle C}}=frac{frac{sinalpha}{sin(90^{circ}-2a)}}{frac{sin2alpha}{sin(90^{circ}-alpha)}}=frac{sinalphacosalpha}{cos2alphasin2alpha}=frac{1}{2cos2alpha}.$$
Id est, $cos2alpha=1,$ which is impossible.
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
If $frac{AD}{DC}=frac{1}{2}$ by the law of sines we obtain:
$$frac{1}{2}=frac{AD}{DC}=frac{frac{AD}{BD}}{frac{DC}{BD}}=frac{frac{sinalpha}{sinmeasuredangle A}}{frac{sin2alpha}{sinmeasuredangle C}}=frac{frac{sinalpha}{sin(90^{circ}-2a)}}{frac{sin2alpha}{sin(90^{circ}-alpha)}}=frac{sinalphacosalpha}{cos2alphasin2alpha}=frac{1}{2cos2alpha}.$$
Id est, $cos2alpha=1,$ which is impossible.
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 4 at 15:18
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
add a comment |
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
Strictly out of interest, could you share the thought process which lead you to this?
$endgroup$
– daedsidog
Jan 4 at 15:22
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
@daedsidog It's law of sines. Which step is not clear?
$endgroup$
– Michael Rozenberg
Jan 4 at 15:25
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
It's perfectly clear, I was asking for clues on how would one even realize to use law of sines in this case.
$endgroup$
– daedsidog
Jan 4 at 15:26
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
$begingroup$
@daedsidog See the second step (the second equlity). It's a preparation to using of the law of sines.
$endgroup$
– Michael Rozenberg
Jan 4 at 15:30
add a comment |
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$begingroup$
@Blue If AC is the diameter, how is $AD/DC = 1/2$? Both AD and DC become radii of the circle and therefore their ratio should be 1.
$endgroup$
– daedsidog
Jan 4 at 15:15
1
$begingroup$
Whoops. .......
$endgroup$
– Blue
Jan 4 at 15:17