Parallel lines & triangles
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Two lines $GE$ and $FD$ meet in $A$. They cut each other at $45$ degrees. Both $G$ and $F$ lie on the circumcircle of square $ALBK$ such that $E$ and $D$ lie on the $KB$ and $LB$ respectively without lying on the corners of the square. I'm supposed to prove that $GF$ and $ED$ are parallel.
1) How does one prove parallelism of lines in general?
2) $45$ degrees seems rather specific, could the same be proven with any angle? (i.e. $20$ degrees)
triangle
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locked by quid♦ Dec 11 '18 at 21:04
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$begingroup$
Two lines $GE$ and $FD$ meet in $A$. They cut each other at $45$ degrees. Both $G$ and $F$ lie on the circumcircle of square $ALBK$ such that $E$ and $D$ lie on the $KB$ and $LB$ respectively without lying on the corners of the square. I'm supposed to prove that $GF$ and $ED$ are parallel.
1) How does one prove parallelism of lines in general?
2) $45$ degrees seems rather specific, could the same be proven with any angle? (i.e. $20$ degrees)
triangle
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locked by quid♦ Dec 11 '18 at 21:04
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
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Are they parallel? Take the particular case where $D=B$; then $F=B$, too, so that (still distinct) lines $GF$ and $EB$ meet.
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– Blue
Dec 11 '18 at 8:58
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Yes, the are parallel. However I have no clue how to show this.
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– user574941
Dec 11 '18 at 10:07
2
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Now that you've changed the relevant angle from $30^circ$ to $45^circ$, yes: the lines are parallel. (My counterexample fails, because, when $D=B=F$, we have $G=E=K$, so that the lines are no longer distinct; their meeting at $B$ is no longer problematic. In any case, the counterexample answers your second question.)
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– Blue
Dec 11 '18 at 10:27
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I'm sorry for this inconvenience.
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– user574941
Dec 11 '18 at 10:37
1
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This question belongs an ongoing competition! (see the translated exercises here: math.meta.stackexchange.com/questions/29448/…)
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– Dr. Mathva
Dec 11 '18 at 19:08
comments disabled on deleted / locked posts / reviews |
show 1 more comment
$begingroup$
Two lines $GE$ and $FD$ meet in $A$. They cut each other at $45$ degrees. Both $G$ and $F$ lie on the circumcircle of square $ALBK$ such that $E$ and $D$ lie on the $KB$ and $LB$ respectively without lying on the corners of the square. I'm supposed to prove that $GF$ and $ED$ are parallel.
1) How does one prove parallelism of lines in general?
2) $45$ degrees seems rather specific, could the same be proven with any angle? (i.e. $20$ degrees)
triangle
$endgroup$
Two lines $GE$ and $FD$ meet in $A$. They cut each other at $45$ degrees. Both $G$ and $F$ lie on the circumcircle of square $ALBK$ such that $E$ and $D$ lie on the $KB$ and $LB$ respectively without lying on the corners of the square. I'm supposed to prove that $GF$ and $ED$ are parallel.
1) How does one prove parallelism of lines in general?
2) $45$ degrees seems rather specific, could the same be proven with any angle? (i.e. $20$ degrees)
triangle
triangle
edited Dec 11 '18 at 10:20
asked Dec 11 '18 at 7:07
user574941
locked by quid♦ Dec 11 '18 at 21:04
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
locked by quid♦ Dec 11 '18 at 21:04
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
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Are they parallel? Take the particular case where $D=B$; then $F=B$, too, so that (still distinct) lines $GF$ and $EB$ meet.
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– Blue
Dec 11 '18 at 8:58
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Yes, the are parallel. However I have no clue how to show this.
$endgroup$
– user574941
Dec 11 '18 at 10:07
2
$begingroup$
Now that you've changed the relevant angle from $30^circ$ to $45^circ$, yes: the lines are parallel. (My counterexample fails, because, when $D=B=F$, we have $G=E=K$, so that the lines are no longer distinct; their meeting at $B$ is no longer problematic. In any case, the counterexample answers your second question.)
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– Blue
Dec 11 '18 at 10:27
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I'm sorry for this inconvenience.
$endgroup$
– user574941
Dec 11 '18 at 10:37
1
$begingroup$
This question belongs an ongoing competition! (see the translated exercises here: math.meta.stackexchange.com/questions/29448/…)
$endgroup$
– Dr. Mathva
Dec 11 '18 at 19:08
comments disabled on deleted / locked posts / reviews |
show 1 more comment
$begingroup$
Are they parallel? Take the particular case where $D=B$; then $F=B$, too, so that (still distinct) lines $GF$ and $EB$ meet.
$endgroup$
– Blue
Dec 11 '18 at 8:58
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Yes, the are parallel. However I have no clue how to show this.
$endgroup$
– user574941
Dec 11 '18 at 10:07
2
$begingroup$
Now that you've changed the relevant angle from $30^circ$ to $45^circ$, yes: the lines are parallel. (My counterexample fails, because, when $D=B=F$, we have $G=E=K$, so that the lines are no longer distinct; their meeting at $B$ is no longer problematic. In any case, the counterexample answers your second question.)
$endgroup$
– Blue
Dec 11 '18 at 10:27
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I'm sorry for this inconvenience.
$endgroup$
– user574941
Dec 11 '18 at 10:37
1
$begingroup$
This question belongs an ongoing competition! (see the translated exercises here: math.meta.stackexchange.com/questions/29448/…)
$endgroup$
– Dr. Mathva
Dec 11 '18 at 19:08
$begingroup$
Are they parallel? Take the particular case where $D=B$; then $F=B$, too, so that (still distinct) lines $GF$ and $EB$ meet.
$endgroup$
– Blue
Dec 11 '18 at 8:58
$begingroup$
Are they parallel? Take the particular case where $D=B$; then $F=B$, too, so that (still distinct) lines $GF$ and $EB$ meet.
$endgroup$
– Blue
Dec 11 '18 at 8:58
$begingroup$
Yes, the are parallel. However I have no clue how to show this.
$endgroup$
– user574941
Dec 11 '18 at 10:07
$begingroup$
Yes, the are parallel. However I have no clue how to show this.
$endgroup$
– user574941
Dec 11 '18 at 10:07
2
2
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Now that you've changed the relevant angle from $30^circ$ to $45^circ$, yes: the lines are parallel. (My counterexample fails, because, when $D=B=F$, we have $G=E=K$, so that the lines are no longer distinct; their meeting at $B$ is no longer problematic. In any case, the counterexample answers your second question.)
$endgroup$
– Blue
Dec 11 '18 at 10:27
$begingroup$
Now that you've changed the relevant angle from $30^circ$ to $45^circ$, yes: the lines are parallel. (My counterexample fails, because, when $D=B=F$, we have $G=E=K$, so that the lines are no longer distinct; their meeting at $B$ is no longer problematic. In any case, the counterexample answers your second question.)
$endgroup$
– Blue
Dec 11 '18 at 10:27
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I'm sorry for this inconvenience.
$endgroup$
– user574941
Dec 11 '18 at 10:37
$begingroup$
I'm sorry for this inconvenience.
$endgroup$
– user574941
Dec 11 '18 at 10:37
1
1
$begingroup$
This question belongs an ongoing competition! (see the translated exercises here: math.meta.stackexchange.com/questions/29448/…)
$endgroup$
– Dr. Mathva
Dec 11 '18 at 19:08
$begingroup$
This question belongs an ongoing competition! (see the translated exercises here: math.meta.stackexchange.com/questions/29448/…)
$endgroup$
– Dr. Mathva
Dec 11 '18 at 19:08
comments disabled on deleted / locked posts / reviews |
show 1 more comment
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Are they parallel? Take the particular case where $D=B$; then $F=B$, too, so that (still distinct) lines $GF$ and $EB$ meet.
$endgroup$
– Blue
Dec 11 '18 at 8:58
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Yes, the are parallel. However I have no clue how to show this.
$endgroup$
– user574941
Dec 11 '18 at 10:07
2
$begingroup$
Now that you've changed the relevant angle from $30^circ$ to $45^circ$, yes: the lines are parallel. (My counterexample fails, because, when $D=B=F$, we have $G=E=K$, so that the lines are no longer distinct; their meeting at $B$ is no longer problematic. In any case, the counterexample answers your second question.)
$endgroup$
– Blue
Dec 11 '18 at 10:27
$begingroup$
I'm sorry for this inconvenience.
$endgroup$
– user574941
Dec 11 '18 at 10:37
1
$begingroup$
This question belongs an ongoing competition! (see the translated exercises here: math.meta.stackexchange.com/questions/29448/…)
$endgroup$
– Dr. Mathva
Dec 11 '18 at 19:08