Geometry involving circumcenters
$begingroup$
In an acute-angled triangle $ABC$, a point $D$ lies on the segment $BC$. Let $O_1$,$O_2$ denote the circumcentres of triangles $ABD$ and $ACD$, respectively. Prove that the line joining the circumcentre of triangle $ABC$ and the orthocentre of triangle $O_1O_2D$ is parallel to $BC$.
Supposing that the circumcentre of $Delta$$ABC$ is $O$, and the orthocenter of $Delta$$O_1O_2D$ is H, I could prove that $A,O_1,O,H,O_2$ lie on a circle. After that I cannot figure out how to do. Please help.
[Any other better solution is also welcome :)]
geometry
asked Jan 1 at 11:26
YellowYellow
16011
$endgroup$
|
show 1 more comment
$begingroup$
In an acute-angled triangle $ABC$, a point $D$ lies on the segment $BC$. Let $O_1$,$O_2$ denote the circumcentres of triangles $ABD$ and $ACD$, respectively. Prove that the line joining the circumcentre of triangle $ABC$ and the orthocentre of triangle $O_1O_2D$ is parallel to $BC$.
Supposing that the circumcentre of $Delta$$ABC$ is $O$, and the orthocenter of $Delta$$O_1O_2D$ is H, I could prove that $A,O_1,O,H,O_2$ lie on a circle. After that I cannot figure out how to do. Please help.
[Any other better solution is also welcome :)]
geometry
asked Jan 1 at 11:26
YellowYellow
16011
$endgroup$
$begingroup$
Please add a diagram.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:29
$begingroup$
Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations.
$endgroup$
– Yellow
Jan 1 at 16:50
$begingroup$
This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:55
$begingroup$
I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!
$endgroup$
– Yellow
Jan 1 at 17:02
$begingroup$
You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P).
$endgroup$
– Anubhab Ghosal
Jan 1 at 19:23
|
show 1 more comment
$begingroup$
In an acute-angled triangle $ABC$, a point $D$ lies on the segment $BC$. Let $O_1$,$O_2$ denote the circumcentres of triangles $ABD$ and $ACD$, respectively. Prove that the line joining the circumcentre of triangle $ABC$ and the orthocentre of triangle $O_1O_2D$ is parallel to $BC$.
Supposing that the circumcentre of $Delta$$ABC$ is $O$, and the orthocenter of $Delta$$O_1O_2D$ is H, I could prove that $A,O_1,O,H,O_2$ lie on a circle. After that I cannot figure out how to do. Please help.
[Any other better solution is also welcome :)]
geometry
asked Jan 1 at 11:26
YellowYellow
16011
$endgroup$
In an acute-angled triangle $ABC$, a point $D$ lies on the segment $BC$. Let $O_1$,$O_2$ denote the circumcentres of triangles $ABD$ and $ACD$, respectively. Prove that the line joining the circumcentre of triangle $ABC$ and the orthocentre of triangle $O_1O_2D$ is parallel to $BC$.
Supposing that the circumcentre of $Delta$$ABC$ is $O$, and the orthocenter of $Delta$$O_1O_2D$ is H, I could prove that $A,O_1,O,H,O_2$ lie on a circle. After that I cannot figure out how to do. Please help.
[Any other better solution is also welcome :)]
geometry
geometry
asked Jan 1 at 11:26
YellowYellow
16011
asked Jan 1 at 11:26
YellowYellow
16011
edited Jan 1 at 11:55
Yellow
asked Jan 1 at 11:26
YellowYellow
16011
asked Jan 1 at 11:26
YellowYellow
16011
asked Jan 1 at 11:26
YellowYellow
16011
16011
$begingroup$
Please add a diagram.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:29
$begingroup$
Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations.
$endgroup$
– Yellow
Jan 1 at 16:50
$begingroup$
This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:55
$begingroup$
I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!
$endgroup$
– Yellow
Jan 1 at 17:02
$begingroup$
You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P).
$endgroup$
– Anubhab Ghosal
Jan 1 at 19:23
|
show 1 more comment
$begingroup$
Please add a diagram.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:29
$begingroup$
Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations.
$endgroup$
– Yellow
Jan 1 at 16:50
$begingroup$
This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:55
$begingroup$
I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!
$endgroup$
– Yellow
Jan 1 at 17:02
$begingroup$
You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P).
$endgroup$
– Anubhab Ghosal
Jan 1 at 19:23
$begingroup$
Please add a diagram.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:29
$begingroup$
Please add a diagram.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:29
$begingroup$
Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations.
$endgroup$
– Yellow
Jan 1 at 16:50
$begingroup$
Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations.
$endgroup$
– Yellow
Jan 1 at 16:50
$begingroup$
This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:55
$begingroup$
This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:55
$begingroup$
I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!
$endgroup$
– Yellow
Jan 1 at 17:02
$begingroup$
I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!
$endgroup$
– Yellow
Jan 1 at 17:02
$begingroup$
You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P).
$endgroup$
– Anubhab Ghosal
Jan 1 at 19:23
$begingroup$
You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P).
$endgroup$
– Anubhab Ghosal
Jan 1 at 19:23
|
show 1 more comment
1 Answer
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$begingroup$
I proceded starting from your suggestions and then by the following path.
You can use your result to state that
begin{equation}
angle O_2OH congangle O_2AH cong angle O_2DH,tag{1}label{eq:cong1}
end{equation}
where the first congruence is due to the fact that the angles are subtended by the same chord $O_2H$, and the second congruence is due to the fact that triangle $AO_2D$ is isosceles. Call then $M_1$ the middle point of $BC$ and $M_2$ the middle point of $AC$. Note that
begin{equation}
angle O_2OM_1 + angle ACB cong pi,tag{2}label{eq:sum1}
end{equation}
which can be obtained by adding up the internal angles of quadrilateral $CM_2OM_1$. Now subtract and add $angle O_2OH$ on the LHS of eqref{eq:sum1}, getting
begin{eqnarray}
angle O_2OM_1 -angle O_2OH+ angle ACB + angle O_2OH&cong& pi\
angle HOM_1 + angle ACB + angle O_2OH cong pi.
end{eqnarray}
Finally, by using eqref{eq:cong1} and properties of the circumcenter $O_2$, show that
begin{equation}
angle ACB + angle O_2OH cong frac{pi}{2},
end{equation}
which will lead to the thesis.
answered Jan 1 at 15:52
MatteoMatteo
34529
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I proceded starting from your suggestions and then by the following path.
You can use your result to state that
begin{equation}
angle O_2OH congangle O_2AH cong angle O_2DH,tag{1}label{eq:cong1}
end{equation}
where the first congruence is due to the fact that the angles are subtended by the same chord $O_2H$, and the second congruence is due to the fact that triangle $AO_2D$ is isosceles. Call then $M_1$ the middle point of $BC$ and $M_2$ the middle point of $AC$. Note that
begin{equation}
angle O_2OM_1 + angle ACB cong pi,tag{2}label{eq:sum1}
end{equation}
which can be obtained by adding up the internal angles of quadrilateral $CM_2OM_1$. Now subtract and add $angle O_2OH$ on the LHS of eqref{eq:sum1}, getting
begin{eqnarray}
angle O_2OM_1 -angle O_2OH+ angle ACB + angle O_2OH&cong& pi\
angle HOM_1 + angle ACB + angle O_2OH cong pi.
end{eqnarray}
Finally, by using eqref{eq:cong1} and properties of the circumcenter $O_2$, show that
begin{equation}
angle ACB + angle O_2OH cong frac{pi}{2},
end{equation}
which will lead to the thesis.
answered Jan 1 at 15:52
MatteoMatteo
34529
$endgroup$
add a comment |
$begingroup$
I proceded starting from your suggestions and then by the following path.
You can use your result to state that
begin{equation}
angle O_2OH congangle O_2AH cong angle O_2DH,tag{1}label{eq:cong1}
end{equation}
where the first congruence is due to the fact that the angles are subtended by the same chord $O_2H$, and the second congruence is due to the fact that triangle $AO_2D$ is isosceles. Call then $M_1$ the middle point of $BC$ and $M_2$ the middle point of $AC$. Note that
begin{equation}
angle O_2OM_1 + angle ACB cong pi,tag{2}label{eq:sum1}
end{equation}
which can be obtained by adding up the internal angles of quadrilateral $CM_2OM_1$. Now subtract and add $angle O_2OH$ on the LHS of eqref{eq:sum1}, getting
begin{eqnarray}
angle O_2OM_1 -angle O_2OH+ angle ACB + angle O_2OH&cong& pi\
angle HOM_1 + angle ACB + angle O_2OH cong pi.
end{eqnarray}
Finally, by using eqref{eq:cong1} and properties of the circumcenter $O_2$, show that
begin{equation}
angle ACB + angle O_2OH cong frac{pi}{2},
end{equation}
which will lead to the thesis.
answered Jan 1 at 15:52
MatteoMatteo
34529
$endgroup$
add a comment |
$begingroup$
I proceded starting from your suggestions and then by the following path.
You can use your result to state that
begin{equation}
angle O_2OH congangle O_2AH cong angle O_2DH,tag{1}label{eq:cong1}
end{equation}
where the first congruence is due to the fact that the angles are subtended by the same chord $O_2H$, and the second congruence is due to the fact that triangle $AO_2D$ is isosceles. Call then $M_1$ the middle point of $BC$ and $M_2$ the middle point of $AC$. Note that
begin{equation}
angle O_2OM_1 + angle ACB cong pi,tag{2}label{eq:sum1}
end{equation}
which can be obtained by adding up the internal angles of quadrilateral $CM_2OM_1$. Now subtract and add $angle O_2OH$ on the LHS of eqref{eq:sum1}, getting
begin{eqnarray}
angle O_2OM_1 -angle O_2OH+ angle ACB + angle O_2OH&cong& pi\
angle HOM_1 + angle ACB + angle O_2OH cong pi.
end{eqnarray}
Finally, by using eqref{eq:cong1} and properties of the circumcenter $O_2$, show that
begin{equation}
angle ACB + angle O_2OH cong frac{pi}{2},
end{equation}
which will lead to the thesis.
answered Jan 1 at 15:52
MatteoMatteo
34529
$endgroup$
I proceded starting from your suggestions and then by the following path.
You can use your result to state that
begin{equation}
angle O_2OH congangle O_2AH cong angle O_2DH,tag{1}label{eq:cong1}
end{equation}
where the first congruence is due to the fact that the angles are subtended by the same chord $O_2H$, and the second congruence is due to the fact that triangle $AO_2D$ is isosceles. Call then $M_1$ the middle point of $BC$ and $M_2$ the middle point of $AC$. Note that
begin{equation}
angle O_2OM_1 + angle ACB cong pi,tag{2}label{eq:sum1}
end{equation}
which can be obtained by adding up the internal angles of quadrilateral $CM_2OM_1$. Now subtract and add $angle O_2OH$ on the LHS of eqref{eq:sum1}, getting
begin{eqnarray}
angle O_2OM_1 -angle O_2OH+ angle ACB + angle O_2OH&cong& pi\
angle HOM_1 + angle ACB + angle O_2OH cong pi.
end{eqnarray}
Finally, by using eqref{eq:cong1} and properties of the circumcenter $O_2$, show that
begin{equation}
angle ACB + angle O_2OH cong frac{pi}{2},
end{equation}
which will lead to the thesis.
answered Jan 1 at 15:52
MatteoMatteo
34529
answered Jan 1 at 15:52
MatteoMatteo
34529
answered Jan 1 at 15:52
MatteoMatteo
34529
answered Jan 1 at 15:52
MatteoMatteo
34529
34529
add a comment |
add a comment |
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$begingroup$
Please add a diagram.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:29
$begingroup$
Well, the original question did not provide a diagram, the reader had to draw the diagram himself/ herself because there may be variations.
$endgroup$
– Yellow
Jan 1 at 16:50
$begingroup$
This is not a site like Brilliant or AOPS. Here, the questioner must show effort and then the answer is to help him/her to complete the solution. One has to draw a diagram to solve the problem. Hence, the question is easier to comprehend with a diagram. As for the fact that multiple configurations are possible, draw any possible configuration.
$endgroup$
– Anubhab Ghosal
Jan 1 at 16:55
$begingroup$
I do understand, @Anubhab Ghosal. I am not simply posting questions or asking doubts. I do write whatever I have done as a part of my attempt to solve the problem. But please see, I registered with this website just yesterday and I am yet to explore and learn more on this website. So maybe in my future posts I will try my best to post a figure, along with the question, too. Anyways, thanks for the suggestions!
$endgroup$
– Yellow
Jan 1 at 17:02
$begingroup$
You are welcome. Welcome to Math.SE. Enjoy! (I am not sure if I am the right person to welcome you as I myself joined only a month back. :P).
$endgroup$
– Anubhab Ghosal
Jan 1 at 19:23