Mixing
Another beauty hidden in a simple triangle (2)
Clash Royale CLAN TAG#URR8PPP
up vote
17
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favorite
Let the inscribed circle of triangle $ABC$ touch its sides $BC, CA, AB$ at $F, G, E$. Denote the center of this circle by $D$, and the midpoint of $BC$ by $M$. Prove that lines $AM$, $EG$ and $DF$ concur.
My idea was to consider the intersection of lines $EG$ and $FD$ first, then to draw a line through intersection point $H$ parallel to $BC$. That line will intersect triangle at some points $I$ and $J$ (not shown). If I could prove that $IH=JH$, that would automatically prove that $AM$ has to pass through point $H$ as well. However, this day was so hot here in my city that I could not think more about this problem. I can only hope that it's more pleasant elsewhere :)
euclidean-geometry triangle
edited Aug 25 at 20:39
darij grinberg
9,32632960
asked Aug 25 at 20:20
Oldboy
2,9221318
add a comment |Â
up vote
17
down vote
favorite
Let the inscribed circle of triangle $ABC$ touch its sides $BC, CA, AB$ at $F, G, E$. Denote the center of this circle by $D$, and the midpoint of $BC$ by $M$. Prove that lines $AM$, $EG$ and $DF$ concur.
My idea was to consider the intersection of lines $EG$ and $FD$ first, then to draw a line through intersection point $H$ parallel to $BC$. That line will intersect triangle at some points $I$ and $J$ (not shown). If I could prove that $IH=JH$, that would automatically prove that $AM$ has to pass through point $H$ as well. However, this day was so hot here in my city that I could not think more about this problem. I can only hope that it's more pleasant elsewhere :)
euclidean-geometry triangle
edited Aug 25 at 20:39
darij grinberg
9,32632960
asked Aug 25 at 20:20
Oldboy
2,9221318
1
The straightforward way is to compute the ratios $sin measuredangle BAH / sin measuredangle HAC$ and $sin measuredangle BAM / sin measuredangle MAC$ (both can be shown to equal $b / c$ using the sine law in triangles $EAH$, $HAG$, $EFH$, $HFG$, $ABM$ and $AMC$), and argue that the equality of these ratios yields that the lines $AH$ and $AM$ are identical. There is likely to be a nicer proof.
– darij grinberg
Aug 25 at 20:40
1
@darijgrinberg Thanks, but the problem comes from the geometry book that does not cover trigonometry at all. There has to be some other solution.
– Oldboy
Aug 25 at 20:43
1
The trilinear (or barycentric) coordinates of $E,F,G,D,M$ are straightforward to compute, so they are the equations of $EG,AM,DF$. The given concurrency boils down to checking that a simple determinant is zero.
– Jack D'Aurizio♦
Aug 25 at 20:44
@Oldboy Um, I'm so sorry for you not to adopt any of my proofs...
– mengdie1982
Aug 25 at 22:56
add a comment |Â
up vote
17
down vote
favorite
up vote
17
down vote
favorite
Let the inscribed circle of triangle $ABC$ touch its sides $BC, CA, AB$ at $F, G, E$. Denote the center of this circle by $D$, and the midpoint of $BC$ by $M$. Prove that lines $AM$, $EG$ and $DF$ concur.
My idea was to consider the intersection of lines $EG$ and $FD$ first, then to draw a line through intersection point $H$ parallel to $BC$. That line will intersect triangle at some points $I$ and $J$ (not shown). If I could prove that $IH=JH$, that would automatically prove that $AM$ has to pass through point $H$ as well. However, this day was so hot here in my city that I could not think more about this problem. I can only hope that it's more pleasant elsewhere :)
euclidean-geometry triangle
edited Aug 25 at 20:39
darij grinberg
9,32632960
asked Aug 25 at 20:20
Oldboy
2,9221318
Let the inscribed circle of triangle $ABC$ touch its sides $BC, CA, AB$ at $F, G, E$. Denote the center of this circle by $D$, and the midpoint of $BC$ by $M$. Prove that lines $AM$, $EG$ and $DF$ concur.
My idea was to consider the intersection of lines $EG$ and $FD$ first, then to draw a line through intersection point $H$ parallel to $BC$. That line will intersect triangle at some points $I$ and $J$ (not shown). If I could prove that $IH=JH$, that would automatically prove that $AM$ has to pass through point $H$ as well. However, this day was so hot here in my city that I could not think more about this problem. I can only hope that it's more pleasant elsewhere :)
euclidean-geometry triangle
edited Aug 25 at 20:39
darij grinberg
9,32632960
asked Aug 25 at 20:20
Oldboy
2,9221318
edited Aug 25 at 20:39
darij grinberg
9,32632960
edited Aug 25 at 20:39
darij grinberg
9,32632960
edited Aug 25 at 20:39
darij grinberg
9,32632960
9,32632960
asked Aug 25 at 20:20
Oldboy
2,9221318
asked Aug 25 at 20:20
Oldboy
2,9221318
asked Aug 25 at 20:20
Oldboy
2,9221318
2,9221318
1
The straightforward way is to compute the ratios $sin measuredangle BAH / sin measuredangle HAC$ and $sin measuredangle BAM / sin measuredangle MAC$ (both can be shown to equal $b / c$ using the sine law in triangles $EAH$, $HAG$, $EFH$, $HFG$, $ABM$ and $AMC$), and argue that the equality of these ratios yields that the lines $AH$ and $AM$ are identical. There is likely to be a nicer proof.
– darij grinberg
Aug 25 at 20:40
1
@darijgrinberg Thanks, but the problem comes from the geometry book that does not cover trigonometry at all. There has to be some other solution.
– Oldboy
Aug 25 at 20:43
1
The trilinear (or barycentric) coordinates of $E,F,G,D,M$ are straightforward to compute, so they are the equations of $EG,AM,DF$. The given concurrency boils down to checking that a simple determinant is zero.
– Jack D'Aurizio♦
Aug 25 at 20:44
@Oldboy Um, I'm so sorry for you not to adopt any of my proofs...
– mengdie1982
Aug 25 at 22:56
add a comment |Â
1
The straightforward way is to compute the ratios $sin measuredangle BAH / sin measuredangle HAC$ and $sin measuredangle BAM / sin measuredangle MAC$ (both can be shown to equal $b / c$ using the sine law in triangles $EAH$, $HAG$, $EFH$, $HFG$, $ABM$ and $AMC$), and argue that the equality of these ratios yields that the lines $AH$ and $AM$ are identical. There is likely to be a nicer proof.
– darij grinberg
Aug 25 at 20:40
1
@darijgrinberg Thanks, but the problem comes from the geometry book that does not cover trigonometry at all. There has to be some other solution.
– Oldboy
Aug 25 at 20:43
1
The trilinear (or barycentric) coordinates of $E,F,G,D,M$ are straightforward to compute, so they are the equations of $EG,AM,DF$. The given concurrency boils down to checking that a simple determinant is zero.
– Jack D'Aurizio♦
Aug 25 at 20:44
@Oldboy Um, I'm so sorry for you not to adopt any of my proofs...
– mengdie1982
Aug 25 at 22:56
1
1
The straightforward way is to compute the ratios $sin measuredangle BAH / sin measuredangle HAC$ and $sin measuredangle BAM / sin measuredangle MAC$ (both can be shown to equal $b / c$ using the sine law in triangles $EAH$, $HAG$, $EFH$, $HFG$, $ABM$ and $AMC$), and argue that the equality of these ratios yields that the lines $AH$ and $AM$ are identical. There is likely to be a nicer proof.
– darij grinberg
Aug 25 at 20:40
The straightforward way is to compute the ratios $sin measuredangle BAH / sin measuredangle HAC$ and $sin measuredangle BAM / sin measuredangle MAC$ (both can be shown to equal $b / c$ using the sine law in triangles $EAH$, $HAG$, $EFH$, $HFG$, $ABM$ and $AMC$), and argue that the equality of these ratios yields that the lines $AH$ and $AM$ are identical. There is likely to be a nicer proof.
– darij grinberg
Aug 25 at 20:40
1
1
@darijgrinberg Thanks, but the problem comes from the geometry book that does not cover trigonometry at all. There has to be some other solution.
– Oldboy
Aug 25 at 20:43
@darijgrinberg Thanks, but the problem comes from the geometry book that does not cover trigonometry at all. There has to be some other solution.
– Oldboy
Aug 25 at 20:43
1
1
The trilinear (or barycentric) coordinates of $E,F,G,D,M$ are straightforward to compute, so they are the equations of $EG,AM,DF$. The given concurrency boils down to checking that a simple determinant is zero.
– Jack D'Aurizio♦
Aug 25 at 20:44
The trilinear (or barycentric) coordinates of $E,F,G,D,M$ are straightforward to compute, so they are the equations of $EG,AM,DF$. The given concurrency boils down to checking that a simple determinant is zero.
– Jack D'Aurizio♦
Aug 25 at 20:44
@Oldboy Um, I'm so sorry for you not to adopt any of my proofs...
– mengdie1982
Aug 25 at 22:56
@Oldboy Um, I'm so sorry for you not to adopt any of my proofs...
– mengdie1982
Aug 25 at 22:56
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
You had the right idea. From $H$ draw a line parallel to $BC$ and call its intersections with $AB$ and $AC$ in order $I$ and $J$. Note that the quadrilaterals $IHDF$ and $HGJD$ are inscribed since $angle DGJ=angle DHJ$ and $angle IHD=angle IFD$ are all right angles. It follows that $angle GDJ=angle GHJ=angle IHF=angle IDF$. So the two triangles $triangle DFI$ and $triangle DGJ$ are congruent for having a leg and a non-right angle congruent. It follows that $DJ$ and $DI$ are congruent and so $HJ$ is congruent to $HI$ and so the line through $A$ and $H$ must go through $M$.
answered Aug 25 at 21:55
Marco
1,55917
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
add a comment |Â
up vote
3
down vote
Proof
Let $AM$ intersect $EG$ at $H$. We are to prove $HF$ is the line of the diameter.
Draw a line $l$ passing through $A$ and parallel to $BC$. Let $EG$ intersect $l$ at $X$.
Thus, we may readily see that $(AB,AC|AM,AX)=-1,$ which shows that the polar line of $H$ with respect to the incircle is $AX$. Hence, $DH perp AX ||BC$, then $F,D,H$ are colinear. The proof is completed so far.
answered Aug 25 at 21:48
mengdie1982
3,643216
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
add a comment |Â
up vote
2
down vote
Another Proof
Denote the the antipodal point of $F$ as $J$. Draw a line $l$ passing through $J$, parallel to $BC$, and intersecting $AB, AC$ at $K, L$ respectively.
Notice that $BCLK$ is a tangential quadrilateral. By Newton's Theorem(Page 156-157) ,which is a degenerated form of Brianchon's theorem, we may obtain $BL,CK,FJ$ are concurrent, namely at $H$. But $KL||BC$, hence $AH$ bisects $BC$ and $KL$. We are done.
answered Aug 25 at 22:31
mengdie1982
3,643216
add a comment |Â
up vote
1
down vote
$IH = JH$ because $M$ is the center point of BC and triangle $AIJ$ is similar with respect to the intersection of its parallel base with line $AM$.
However, you have to show that these two sides are the same due to symmetry about the line $FD$, so the extension of $FD$ is shown to meet the intersection at $H$.
The following diagram achieves this by adding a symmetrical construction about $FD$ with $X = Y$.
answered Aug 26 at 0:23
Phil H
1,9472311
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You had the right idea. From $H$ draw a line parallel to $BC$ and call its intersections with $AB$ and $AC$ in order $I$ and $J$. Note that the quadrilaterals $IHDF$ and $HGJD$ are inscribed since $angle DGJ=angle DHJ$ and $angle IHD=angle IFD$ are all right angles. It follows that $angle GDJ=angle GHJ=angle IHF=angle IDF$. So the two triangles $triangle DFI$ and $triangle DGJ$ are congruent for having a leg and a non-right angle congruent. It follows that $DJ$ and $DI$ are congruent and so $HJ$ is congruent to $HI$ and so the line through $A$ and $H$ must go through $M$.
answered Aug 25 at 21:55
Marco
1,55917
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
add a comment |Â
up vote
3
down vote
accepted
You had the right idea. From $H$ draw a line parallel to $BC$ and call its intersections with $AB$ and $AC$ in order $I$ and $J$. Note that the quadrilaterals $IHDF$ and $HGJD$ are inscribed since $angle DGJ=angle DHJ$ and $angle IHD=angle IFD$ are all right angles. It follows that $angle GDJ=angle GHJ=angle IHF=angle IDF$. So the two triangles $triangle DFI$ and $triangle DGJ$ are congruent for having a leg and a non-right angle congruent. It follows that $DJ$ and $DI$ are congruent and so $HJ$ is congruent to $HI$ and so the line through $A$ and $H$ must go through $M$.
answered Aug 25 at 21:55
Marco
1,55917
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You had the right idea. From $H$ draw a line parallel to $BC$ and call its intersections with $AB$ and $AC$ in order $I$ and $J$. Note that the quadrilaterals $IHDF$ and $HGJD$ are inscribed since $angle DGJ=angle DHJ$ and $angle IHD=angle IFD$ are all right angles. It follows that $angle GDJ=angle GHJ=angle IHF=angle IDF$. So the two triangles $triangle DFI$ and $triangle DGJ$ are congruent for having a leg and a non-right angle congruent. It follows that $DJ$ and $DI$ are congruent and so $HJ$ is congruent to $HI$ and so the line through $A$ and $H$ must go through $M$.
answered Aug 25 at 21:55
Marco
1,55917
You had the right idea. From $H$ draw a line parallel to $BC$ and call its intersections with $AB$ and $AC$ in order $I$ and $J$. Note that the quadrilaterals $IHDF$ and $HGJD$ are inscribed since $angle DGJ=angle DHJ$ and $angle IHD=angle IFD$ are all right angles. It follows that $angle GDJ=angle GHJ=angle IHF=angle IDF$. So the two triangles $triangle DFI$ and $triangle DGJ$ are congruent for having a leg and a non-right angle congruent. It follows that $DJ$ and $DI$ are congruent and so $HJ$ is congruent to $HI$ and so the line through $A$ and $H$ must go through $M$.
answered Aug 25 at 21:55
Marco
1,55917
edited Aug 25 at 22:21
answered Aug 25 at 21:55
Marco
1,55917
answered Aug 25 at 21:55
Marco
1,55917
answered Aug 25 at 21:55
Marco
1,55917
1,55917
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
add a comment |Â
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
I think that you have some typos in your proof but it looks good. The simplest one so far.
– Oldboy
Aug 25 at 22:17
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
@Oldboy, yes I fixed some typos. Thanks.
– Marco
Aug 25 at 22:22
add a comment |Â
up vote
3
down vote
Proof
Let $AM$ intersect $EG$ at $H$. We are to prove $HF$ is the line of the diameter.
Draw a line $l$ passing through $A$ and parallel to $BC$. Let $EG$ intersect $l$ at $X$.
Thus, we may readily see that $(AB,AC|AM,AX)=-1,$ which shows that the polar line of $H$ with respect to the incircle is $AX$. Hence, $DH perp AX ||BC$, then $F,D,H$ are colinear. The proof is completed so far.
answered Aug 25 at 21:48
mengdie1982
3,643216
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
add a comment |Â
up vote
3
down vote
Proof
Let $AM$ intersect $EG$ at $H$. We are to prove $HF$ is the line of the diameter.
Draw a line $l$ passing through $A$ and parallel to $BC$. Let $EG$ intersect $l$ at $X$.
Thus, we may readily see that $(AB,AC|AM,AX)=-1,$ which shows that the polar line of $H$ with respect to the incircle is $AX$. Hence, $DH perp AX ||BC$, then $F,D,H$ are colinear. The proof is completed so far.
answered Aug 25 at 21:48
mengdie1982
3,643216
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Proof
Let $AM$ intersect $EG$ at $H$. We are to prove $HF$ is the line of the diameter.
Draw a line $l$ passing through $A$ and parallel to $BC$. Let $EG$ intersect $l$ at $X$.
Thus, we may readily see that $(AB,AC|AM,AX)=-1,$ which shows that the polar line of $H$ with respect to the incircle is $AX$. Hence, $DH perp AX ||BC$, then $F,D,H$ are colinear. The proof is completed so far.
answered Aug 25 at 21:48
mengdie1982
3,643216
Proof
Let $AM$ intersect $EG$ at $H$. We are to prove $HF$ is the line of the diameter.
Draw a line $l$ passing through $A$ and parallel to $BC$. Let $EG$ intersect $l$ at $X$.
Thus, we may readily see that $(AB,AC|AM,AX)=-1,$ which shows that the polar line of $H$ with respect to the incircle is $AX$. Hence, $DH perp AX ||BC$, then $F,D,H$ are colinear. The proof is completed so far.
answered Aug 25 at 21:48
mengdie1982
3,643216
edited Aug 27 at 15:15
answered Aug 25 at 21:48
mengdie1982
3,643216
answered Aug 25 at 21:48
mengdie1982
3,643216
answered Aug 25 at 21:48
mengdie1982
3,643216
3,643216
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
add a comment |Â
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
Please don't use JPG for non-photographic images.
– Andreas Rejbrand
Aug 26 at 9:41
add a comment |Â
up vote
2
down vote
Another Proof
Denote the the antipodal point of $F$ as $J$. Draw a line $l$ passing through $J$, parallel to $BC$, and intersecting $AB, AC$ at $K, L$ respectively.
Notice that $BCLK$ is a tangential quadrilateral. By Newton's Theorem(Page 156-157) ,which is a degenerated form of Brianchon's theorem, we may obtain $BL,CK,FJ$ are concurrent, namely at $H$. But $KL||BC$, hence $AH$ bisects $BC$ and $KL$. We are done.
answered Aug 25 at 22:31
mengdie1982
3,643216
add a comment |Â
up vote
2
down vote
Another Proof
Denote the the antipodal point of $F$ as $J$. Draw a line $l$ passing through $J$, parallel to $BC$, and intersecting $AB, AC$ at $K, L$ respectively.
Notice that $BCLK$ is a tangential quadrilateral. By Newton's Theorem(Page 156-157) ,which is a degenerated form of Brianchon's theorem, we may obtain $BL,CK,FJ$ are concurrent, namely at $H$. But $KL||BC$, hence $AH$ bisects $BC$ and $KL$. We are done.
answered Aug 25 at 22:31
mengdie1982
3,643216
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Another Proof
Denote the the antipodal point of $F$ as $J$. Draw a line $l$ passing through $J$, parallel to $BC$, and intersecting $AB, AC$ at $K, L$ respectively.
Notice that $BCLK$ is a tangential quadrilateral. By Newton's Theorem(Page 156-157) ,which is a degenerated form of Brianchon's theorem, we may obtain $BL,CK,FJ$ are concurrent, namely at $H$. But $KL||BC$, hence $AH$ bisects $BC$ and $KL$. We are done.
answered Aug 25 at 22:31
mengdie1982
3,643216
Another Proof
Denote the the antipodal point of $F$ as $J$. Draw a line $l$ passing through $J$, parallel to $BC$, and intersecting $AB, AC$ at $K, L$ respectively.
Notice that $BCLK$ is a tangential quadrilateral. By Newton's Theorem(Page 156-157) ,which is a degenerated form of Brianchon's theorem, we may obtain $BL,CK,FJ$ are concurrent, namely at $H$. But $KL||BC$, hence $AH$ bisects $BC$ and $KL$. We are done.
answered Aug 25 at 22:31
mengdie1982
3,643216
edited Aug 25 at 22:51
answered Aug 25 at 22:31
mengdie1982
3,643216
answered Aug 25 at 22:31
mengdie1982
3,643216
answered Aug 25 at 22:31
mengdie1982
3,643216
3,643216
add a comment |Â
add a comment |Â
up vote
1
down vote
$IH = JH$ because $M$ is the center point of BC and triangle $AIJ$ is similar with respect to the intersection of its parallel base with line $AM$.
However, you have to show that these two sides are the same due to symmetry about the line $FD$, so the extension of $FD$ is shown to meet the intersection at $H$.
The following diagram achieves this by adding a symmetrical construction about $FD$ with $X = Y$.
answered Aug 26 at 0:23
Phil H
1,9472311
add a comment |Â
up vote
1
down vote
$IH = JH$ because $M$ is the center point of BC and triangle $AIJ$ is similar with respect to the intersection of its parallel base with line $AM$.
However, you have to show that these two sides are the same due to symmetry about the line $FD$, so the extension of $FD$ is shown to meet the intersection at $H$.
The following diagram achieves this by adding a symmetrical construction about $FD$ with $X = Y$.
answered Aug 26 at 0:23
Phil H
1,9472311
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$IH = JH$ because $M$ is the center point of BC and triangle $AIJ$ is similar with respect to the intersection of its parallel base with line $AM$.
However, you have to show that these two sides are the same due to symmetry about the line $FD$, so the extension of $FD$ is shown to meet the intersection at $H$.
The following diagram achieves this by adding a symmetrical construction about $FD$ with $X = Y$.
answered Aug 26 at 0:23
Phil H
1,9472311
$IH = JH$ because $M$ is the center point of BC and triangle $AIJ$ is similar with respect to the intersection of its parallel base with line $AM$.
However, you have to show that these two sides are the same due to symmetry about the line $FD$, so the extension of $FD$ is shown to meet the intersection at $H$.
The following diagram achieves this by adding a symmetrical construction about $FD$ with $X = Y$.
answered Aug 26 at 0:23
Phil H
1,9472311
answered Aug 26 at 0:23
Phil H
1,9472311
answered Aug 26 at 0:23
Phil H
1,9472311
answered Aug 26 at 0:23
Phil H
1,9472311
1,9472311
add a comment |Â
add a comment |Â
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1
The straightforward way is to compute the ratios $sin measuredangle BAH / sin measuredangle HAC$ and $sin measuredangle BAM / sin measuredangle MAC$ (both can be shown to equal $b / c$ using the sine law in triangles $EAH$, $HAG$, $EFH$, $HFG$, $ABM$ and $AMC$), and argue that the equality of these ratios yields that the lines $AH$ and $AM$ are identical. There is likely to be a nicer proof.
– darij grinberg
Aug 25 at 20:40
1
@darijgrinberg Thanks, but the problem comes from the geometry book that does not cover trigonometry at all. There has to be some other solution.
– Oldboy
Aug 25 at 20:43
1
The trilinear (or barycentric) coordinates of $E,F,G,D,M$ are straightforward to compute, so they are the equations of $EG,AM,DF$. The given concurrency boils down to checking that a simple determinant is zero.
– Jack D'Aurizio♦
Aug 25 at 20:44
@Oldboy Um, I'm so sorry for you not to adopt any of my proofs...
– mengdie1982
Aug 25 at 22:56