Three Altitudes of a triangle are concurrent
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I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
$endgroup$
add a comment |
$begingroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
$endgroup$
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
add a comment |
$begingroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
$endgroup$
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.
I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.
Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.
geometry euclidean-geometry triangles
geometry euclidean-geometry triangles
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
edited Mar 16 '16 at 22:45
Smurf
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
asked Mar 16 '16 at 20:26
SmurfSmurf
593313
593313
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
add a comment |
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22
add a comment |
2 Answers
2
active
oldest
votes
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Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
$endgroup$
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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oldest
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oldest
votes
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
$endgroup$
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
$endgroup$
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
$endgroup$
Since $AEFC$ is cyclic, $angle FEB =angle ACB$.
Since $BEDF$ is cyclic, $angle FEB=angle BDF$.
Thus $angle GDF+angle ACB=180^circ$.
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
edited Mar 20 '16 at 14:46
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
answered Mar 17 '16 at 5:51
Quang HoangQuang Hoang
13.4k1233
13.4k1233
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
The problem I find is to show that the quadrilaterals are cyclic from Euclid's books, since it does not seem to be any proposition about it.
$endgroup$
– Smurf
Mar 17 '16 at 7:53
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
$begingroup$
In this case, it's fairly easy: $A$, $E$, $F$, $C$ all lie on the circle with diameter $AC$, while $B$, $E$, $D$, $F$ lie on the circle with diameter $BD$. In general, you can check whether the two opposite angles add up to $180^circ$ (like in $BEDF$) or the two diagonal-side angles looking at a side are equal (like in $AEFC$.
$endgroup$
– Quang Hoang
Mar 17 '16 at 13:47
1
1
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Two things: first I think $angle BDE$ should read $angle BDF$ in line 2.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:18
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
$begingroup$
Also, one way to justify that $AEFC$ is cyclic is to prove the converse of Prop 31 which says the angle in the semicircle is right. Show that a right angle standing on a straight line has its vertex fall on the semicircle described on that straight line as a diameter. In your diagram, we notice that $angle AEC$ and $angle AFC$ are right by construction. This would be sufficient to show that $AEFC$ is a cyclic quadrilateral.
$endgroup$
– Eric Haney
Mar 20 '16 at 6:27
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
$begingroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
$endgroup$
The easiest way I know of showing the altitudes of $ABC$ are concurrent is (1)Prove the right bisectors of a triangle are concurrent. (2) Draw line $l_A$ thru $A$ parallel to $BC$, line $l_B$ thru $B$ parallel to $CA,$ and $l_C$ thru $C$ parallel to $AB.$ These lines meet pair-wise at points $A',B',C'.$ The altitudes of $ABC$ are the right bisectors of $A'B'C'.$
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
answered Mar 16 '16 at 22:33
DanielWainfleetDanielWainfleet
36k31648
36k31648
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
$begingroup$
There is a lot to argument there. I am trying to do a step by step proof with explicit mention of the propositions used on each step. Also would like to follow the cyclic quadrilaterals path.
$endgroup$
– Smurf
Mar 16 '16 at 22:42
add a comment |
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$begingroup$
I think you mean AF and GCFD instead of AE and GCED. (The F looks like it could be an E in the diagram.)
$endgroup$
– Théophile
Mar 16 '16 at 22:38
$begingroup$
This doesn't help you, but you could do this just using formulas if you were allowed.
$endgroup$
– barrycarter
Mar 16 '16 at 23:22