Proving colinearity of 3 points(basic Euclidean geometry)
1
$begingroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
$endgroup$
add a comment |
1
$begingroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
$endgroup$
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
add a comment |
1
1
1
$begingroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
$endgroup$
In the above image, AP is bisector of the angle BAC and DP is a perpendicular bisector of the segment BC.
How can it be proved that the points E, D, F are colinear?
euclidean-geometry
euclidean-geometry
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
asked Dec 3 '18 at 10:02
Ki Yoon EumKi Yoon Eum
174
174
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
add a comment |
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
1
1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32
add a comment |
1 Answer
1
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oldest
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0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
0
0
0
$begingroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
$endgroup$
Note that $P$ is the mid-point of the minor arc BC of the circumcircle ABC. So EDF is the Simson line of the point $P$.
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
answered Dec 4 '18 at 10:52
user10354138user10354138
7,3772925
7,3772925
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
So, how is the Simpson "line" a line?
$endgroup$
– Oscar Lanzi
Dec 4 '18 at 13:55
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
$begingroup$
The linked wiki article has a proof, as does many other sites such as cut-the-knot. I don't think I should regurgitate it here.
$endgroup$
– user10354138
Dec 4 '18 at 13:57
add a comment |
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1
$begingroup$
You cannot in general ! because it is true only for some special position of $BC$
$endgroup$
– Emilio Novati
Dec 3 '18 at 10:29
$begingroup$
What special position? This is an exercise problem in "Euclidean and non-Euclidean geometry" and there is no such mention
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 11:23
1
$begingroup$
And... I solve it actually using Menelaus' theorem!
$endgroup$
– Ki Yoon Eum
Dec 3 '18 at 12:23
$begingroup$
Can you add your solution (as an answer)?
$endgroup$
– Berci
Dec 3 '18 at 19:32