Show that $OX=OY$.
$begingroup$
Let $O$ the circumcenter of an acute triangle $ABC$. Let $alpha $ the circle trough $A $ and $B $ tangent to $[AC] $, and $beta $ the circle trough $A, C $ tangent to $[AB] $. A line trough $A $ intersects the second time the circles $alpha $ and $beta $ in $X $ and $Y $. Show that $OX=OY $.
I try to check that the second point of intersection of the circles is $O $. With GeoGebra this fact is false.
geometry euclidean-geometry triangle circle geometric-transformation
edited Dec 29 '18 at 15:16
greedoid
44k1155109
asked Nov 7 '18 at 14:10
rafarafa
606212
$endgroup$
add a comment |
$begingroup$
Let $O$ the circumcenter of an acute triangle $ABC$. Let $alpha $ the circle trough $A $ and $B $ tangent to $[AC] $, and $beta $ the circle trough $A, C $ tangent to $[AB] $. A line trough $A $ intersects the second time the circles $alpha $ and $beta $ in $X $ and $Y $. Show that $OX=OY $.
I try to check that the second point of intersection of the circles is $O $. With GeoGebra this fact is false.
geometry euclidean-geometry triangle circle geometric-transformation
edited Dec 29 '18 at 15:16
greedoid
44k1155109
asked Nov 7 '18 at 14:10
rafarafa
606212
$endgroup$
3
$begingroup$
If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing.
$endgroup$
– Quang Hoang
Nov 7 '18 at 14:25
add a comment |
$begingroup$
Let $O$ the circumcenter of an acute triangle $ABC$. Let $alpha $ the circle trough $A $ and $B $ tangent to $[AC] $, and $beta $ the circle trough $A, C $ tangent to $[AB] $. A line trough $A $ intersects the second time the circles $alpha $ and $beta $ in $X $ and $Y $. Show that $OX=OY $.
I try to check that the second point of intersection of the circles is $O $. With GeoGebra this fact is false.
geometry euclidean-geometry triangle circle geometric-transformation
edited Dec 29 '18 at 15:16
greedoid
44k1155109
asked Nov 7 '18 at 14:10
rafarafa
606212
$endgroup$
Let $O$ the circumcenter of an acute triangle $ABC$. Let $alpha $ the circle trough $A $ and $B $ tangent to $[AC] $, and $beta $ the circle trough $A, C $ tangent to $[AB] $. A line trough $A $ intersects the second time the circles $alpha $ and $beta $ in $X $ and $Y $. Show that $OX=OY $.
I try to check that the second point of intersection of the circles is $O $. With GeoGebra this fact is false.
geometry euclidean-geometry triangle circle geometric-transformation
geometry euclidean-geometry triangle circle geometric-transformation
edited Dec 29 '18 at 15:16
greedoid
44k1155109
asked Nov 7 '18 at 14:10
rafarafa
606212
edited Dec 29 '18 at 15:16
greedoid
44k1155109
asked Nov 7 '18 at 14:10
rafarafa
606212
edited Dec 29 '18 at 15:16
greedoid
44k1155109
edited Dec 29 '18 at 15:16
greedoid
44k1155109
edited Dec 29 '18 at 15:16
greedoid
44k1155109
44k1155109
asked Nov 7 '18 at 14:10
rafarafa
606212
asked Nov 7 '18 at 14:10
rafarafa
606212
asked Nov 7 '18 at 14:10
rafarafa
606212
606212
3
$begingroup$
If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing.
$endgroup$
– Quang Hoang
Nov 7 '18 at 14:25
add a comment |
3
$begingroup$
If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing.
$endgroup$
– Quang Hoang
Nov 7 '18 at 14:25
3
3
$begingroup$
If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing.
$endgroup$
– Quang Hoang
Nov 7 '18 at 14:25
$begingroup$
If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing.
$endgroup$
– Quang Hoang
Nov 7 '18 at 14:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since I’m on the phone and can’t type properly I will make some hints:
- Show that triangles $ABX$ and $ACY$ are similar.
- Show that $AEOD$ is a parallelogram.
- Show that triangles $XEO$ and $ODY$ are congruent.
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
add a comment |
$begingroup$
My proof does not use similarity. So it's not just a copy of Quang's (excellent) idea.
Notice that:
$$ODbot AB,space EAbot AB implies DOparallel AE$$
$$OEbot AC,space DAbot AC implies DAparallel OE$$
Obviously, quadrilateral ADOE is a parallelogram (opposite sides are parallel) and because of that:
$$DX=DA=OEtag{1}$$
$$DO=AE=EYtag{2}$$
Introduce angles $delta=angle EAY$, $gamma=angle BAC$
$$angle AEY=180^circ-2delta$$
$$angle AEO=gamma$$
(the last is true because angles ∠AEO and ∠BAC have perpendicular legs)
$$angle OEY=360^circ-angle AEY -angle AEO=180^circ+2delta-gammatag{3}$$
On the other side opposite angles in parallelogram $angle DAE$ and $angle DOE$ are equal and $angle DOE$ is obtuse angle with perpendicular legs with respect to angle $angle BAC=gamma$:
$$angle DAE=angle DOE=180^circ-gamma$$
$$angle DAX=180^circ-angle DAE-angle EAY=gamma-delta$$
$$angle DXA=angle DAX=gamma-delta$$
$$angle XDA=180^circ-angle DAX-angle DXA=180^circ-2gamma+2delta$$
$$angle ADO=gamma$$
(the last is true because angles $angle ADO$ and $angle BAC$ have perpendicular legs)
Finally:
$$angle XDO=angle XDA+angle ADO=180^circ+2delta-gammatag{4}$$
From (3) and (4):
$$angle XDO=angle OEYtag{5}$$
Because of (1), (2) and (5) triangles $triangle XDO$ and $triangle OEY$ are congruent by SAS so it must be that $OX=OY$.
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
2
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votes
$begingroup$
Since I’m on the phone and can’t type properly I will make some hints:
- Show that triangles $ABX$ and $ACY$ are similar.
- Show that $AEOD$ is a parallelogram.
- Show that triangles $XEO$ and $ODY$ are congruent.
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
add a comment |
$begingroup$
Since I’m on the phone and can’t type properly I will make some hints:
- Show that triangles $ABX$ and $ACY$ are similar.
- Show that $AEOD$ is a parallelogram.
- Show that triangles $XEO$ and $ODY$ are congruent.
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
add a comment |
$begingroup$
Since I’m on the phone and can’t type properly I will make some hints:
- Show that triangles $ABX$ and $ACY$ are similar.
- Show that $AEOD$ is a parallelogram.
- Show that triangles $XEO$ and $ODY$ are congruent.
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
$endgroup$
Since I’m on the phone and can’t type properly I will make some hints:
- Show that triangles $ABX$ and $ACY$ are similar.
- Show that $AEOD$ is a parallelogram.
- Show that triangles $XEO$ and $ODY$ are congruent.
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
answered Nov 7 '18 at 14:46
Quang HoangQuang Hoang
13.2k1233
13.2k1233
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
add a comment |
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
Very educational answer !
$endgroup$
– Jean Marie
Nov 7 '18 at 14:53
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@QuangHoang Can you explain why $triangle ABX$ ~ $triangle CBY $? I can deduce from the second assertion which is very clear that $angle AXB=angle CBY $. I need to prove another congruence.
$endgroup$
– rafa
Nov 9 '18 at 6:08
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
$begingroup$
@rafa since $AB$ is a tangent line, $angle BAY = 1/2 arc{ACY} = pi - angle ACY$. So $angle XAB = angle ACY$. Similarly, $angle BXA = angle CAY$.
$endgroup$
– Quang Hoang
Nov 9 '18 at 6:13
add a comment |
$begingroup$
My proof does not use similarity. So it's not just a copy of Quang's (excellent) idea.
Notice that:
$$ODbot AB,space EAbot AB implies DOparallel AE$$
$$OEbot AC,space DAbot AC implies DAparallel OE$$
Obviously, quadrilateral ADOE is a parallelogram (opposite sides are parallel) and because of that:
$$DX=DA=OEtag{1}$$
$$DO=AE=EYtag{2}$$
Introduce angles $delta=angle EAY$, $gamma=angle BAC$
$$angle AEY=180^circ-2delta$$
$$angle AEO=gamma$$
(the last is true because angles ∠AEO and ∠BAC have perpendicular legs)
$$angle OEY=360^circ-angle AEY -angle AEO=180^circ+2delta-gammatag{3}$$
On the other side opposite angles in parallelogram $angle DAE$ and $angle DOE$ are equal and $angle DOE$ is obtuse angle with perpendicular legs with respect to angle $angle BAC=gamma$:
$$angle DAE=angle DOE=180^circ-gamma$$
$$angle DAX=180^circ-angle DAE-angle EAY=gamma-delta$$
$$angle DXA=angle DAX=gamma-delta$$
$$angle XDA=180^circ-angle DAX-angle DXA=180^circ-2gamma+2delta$$
$$angle ADO=gamma$$
(the last is true because angles $angle ADO$ and $angle BAC$ have perpendicular legs)
Finally:
$$angle XDO=angle XDA+angle ADO=180^circ+2delta-gammatag{4}$$
From (3) and (4):
$$angle XDO=angle OEYtag{5}$$
Because of (1), (2) and (5) triangles $triangle XDO$ and $triangle OEY$ are congruent by SAS so it must be that $OX=OY$.
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
$endgroup$
add a comment |
$begingroup$
My proof does not use similarity. So it's not just a copy of Quang's (excellent) idea.
Notice that:
$$ODbot AB,space EAbot AB implies DOparallel AE$$
$$OEbot AC,space DAbot AC implies DAparallel OE$$
Obviously, quadrilateral ADOE is a parallelogram (opposite sides are parallel) and because of that:
$$DX=DA=OEtag{1}$$
$$DO=AE=EYtag{2}$$
Introduce angles $delta=angle EAY$, $gamma=angle BAC$
$$angle AEY=180^circ-2delta$$
$$angle AEO=gamma$$
(the last is true because angles ∠AEO and ∠BAC have perpendicular legs)
$$angle OEY=360^circ-angle AEY -angle AEO=180^circ+2delta-gammatag{3}$$
On the other side opposite angles in parallelogram $angle DAE$ and $angle DOE$ are equal and $angle DOE$ is obtuse angle with perpendicular legs with respect to angle $angle BAC=gamma$:
$$angle DAE=angle DOE=180^circ-gamma$$
$$angle DAX=180^circ-angle DAE-angle EAY=gamma-delta$$
$$angle DXA=angle DAX=gamma-delta$$
$$angle XDA=180^circ-angle DAX-angle DXA=180^circ-2gamma+2delta$$
$$angle ADO=gamma$$
(the last is true because angles $angle ADO$ and $angle BAC$ have perpendicular legs)
Finally:
$$angle XDO=angle XDA+angle ADO=180^circ+2delta-gammatag{4}$$
From (3) and (4):
$$angle XDO=angle OEYtag{5}$$
Because of (1), (2) and (5) triangles $triangle XDO$ and $triangle OEY$ are congruent by SAS so it must be that $OX=OY$.
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
$endgroup$
add a comment |
$begingroup$
My proof does not use similarity. So it's not just a copy of Quang's (excellent) idea.
Notice that:
$$ODbot AB,space EAbot AB implies DOparallel AE$$
$$OEbot AC,space DAbot AC implies DAparallel OE$$
Obviously, quadrilateral ADOE is a parallelogram (opposite sides are parallel) and because of that:
$$DX=DA=OEtag{1}$$
$$DO=AE=EYtag{2}$$
Introduce angles $delta=angle EAY$, $gamma=angle BAC$
$$angle AEY=180^circ-2delta$$
$$angle AEO=gamma$$
(the last is true because angles ∠AEO and ∠BAC have perpendicular legs)
$$angle OEY=360^circ-angle AEY -angle AEO=180^circ+2delta-gammatag{3}$$
On the other side opposite angles in parallelogram $angle DAE$ and $angle DOE$ are equal and $angle DOE$ is obtuse angle with perpendicular legs with respect to angle $angle BAC=gamma$:
$$angle DAE=angle DOE=180^circ-gamma$$
$$angle DAX=180^circ-angle DAE-angle EAY=gamma-delta$$
$$angle DXA=angle DAX=gamma-delta$$
$$angle XDA=180^circ-angle DAX-angle DXA=180^circ-2gamma+2delta$$
$$angle ADO=gamma$$
(the last is true because angles $angle ADO$ and $angle BAC$ have perpendicular legs)
Finally:
$$angle XDO=angle XDA+angle ADO=180^circ+2delta-gammatag{4}$$
From (3) and (4):
$$angle XDO=angle OEYtag{5}$$
Because of (1), (2) and (5) triangles $triangle XDO$ and $triangle OEY$ are congruent by SAS so it must be that $OX=OY$.
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
$endgroup$
My proof does not use similarity. So it's not just a copy of Quang's (excellent) idea.
Notice that:
$$ODbot AB,space EAbot AB implies DOparallel AE$$
$$OEbot AC,space DAbot AC implies DAparallel OE$$
Obviously, quadrilateral ADOE is a parallelogram (opposite sides are parallel) and because of that:
$$DX=DA=OEtag{1}$$
$$DO=AE=EYtag{2}$$
Introduce angles $delta=angle EAY$, $gamma=angle BAC$
$$angle AEY=180^circ-2delta$$
$$angle AEO=gamma$$
(the last is true because angles ∠AEO and ∠BAC have perpendicular legs)
$$angle OEY=360^circ-angle AEY -angle AEO=180^circ+2delta-gammatag{3}$$
On the other side opposite angles in parallelogram $angle DAE$ and $angle DOE$ are equal and $angle DOE$ is obtuse angle with perpendicular legs with respect to angle $angle BAC=gamma$:
$$angle DAE=angle DOE=180^circ-gamma$$
$$angle DAX=180^circ-angle DAE-angle EAY=gamma-delta$$
$$angle DXA=angle DAX=gamma-delta$$
$$angle XDA=180^circ-angle DAX-angle DXA=180^circ-2gamma+2delta$$
$$angle ADO=gamma$$
(the last is true because angles $angle ADO$ and $angle BAC$ have perpendicular legs)
Finally:
$$angle XDO=angle XDA+angle ADO=180^circ+2delta-gammatag{4}$$
From (3) and (4):
$$angle XDO=angle OEYtag{5}$$
Because of (1), (2) and (5) triangles $triangle XDO$ and $triangle OEY$ are congruent by SAS so it must be that $OX=OY$.
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
edited Nov 7 '18 at 15:59
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
answered Nov 7 '18 at 15:53
OldboyOldboy
8,4521936
8,4521936
add a comment |
add a comment |
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3
$begingroup$
If you already have a geogebra drawing, why don’t you include it in the post. It would be more helpful and appealing.
$endgroup$
– Quang Hoang
Nov 7 '18 at 14:25