Proove Line Circumscribed Circle; Incircle; Excircle
up vote
1
down vote
favorite
How can I proove, that the circumscribed circle of a triangle does exactly cross the middle of the line that goes from the incenter of the incircle of the triangle to the excenter of the excircle of the triangle?
geometry proof-writing euclidean-geometry triangle circle
edited Nov 13 at 22:45
Batominovski
31.3k23187
asked Nov 13 at 22:41
calculatormathematical
389
add a comment |
up vote
1
down vote
favorite
How can I proove, that the circumscribed circle of a triangle does exactly cross the middle of the line that goes from the incenter of the incircle of the triangle to the excenter of the excircle of the triangle?
geometry proof-writing euclidean-geometry triangle circle
edited Nov 13 at 22:45
Batominovski
31.3k23187
asked Nov 13 at 22:41
calculatormathematical
389
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How can I proove, that the circumscribed circle of a triangle does exactly cross the middle of the line that goes from the incenter of the incircle of the triangle to the excenter of the excircle of the triangle?
geometry proof-writing euclidean-geometry triangle circle
edited Nov 13 at 22:45
Batominovski
31.3k23187
asked Nov 13 at 22:41
calculatormathematical
389
How can I proove, that the circumscribed circle of a triangle does exactly cross the middle of the line that goes from the incenter of the incircle of the triangle to the excenter of the excircle of the triangle?
geometry proof-writing euclidean-geometry triangle circle
geometry proof-writing euclidean-geometry triangle circle
edited Nov 13 at 22:45
Batominovski
31.3k23187
asked Nov 13 at 22:41
calculatormathematical
389
edited Nov 13 at 22:45
Batominovski
31.3k23187
asked Nov 13 at 22:41
calculatormathematical
389
edited Nov 13 at 22:45
Batominovski
31.3k23187
edited Nov 13 at 22:45
Batominovski
31.3k23187
edited Nov 13 at 22:45
Batominovski
31.3k23187
31.3k23187
asked Nov 13 at 22:41
calculatormathematical
389
asked Nov 13 at 22:41
calculatormathematical
389
asked Nov 13 at 22:41
calculatormathematical
389
389
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
By angle chasing, you can prove the statement. Here are some steps to do so.
Let $ABC$ be a triangle with circumcircle $Gamma$. Write $I$ and $I_a$, respectively, for the incenter and the excenter opposite to the vertex $A$ of the triangle $ABC$. Denote by $M_a$ the midpoint of the arc $BC$ of $Gamma$ that does not contain $A$.
Show that $angle IBI_a=dfrac{pi}{2}=angle ICI_a$. As a consequence, $IBI_aC$ is a cyclic quadrilateral. If $omega_a$ is the circumcircle of the quadrilateral $IBI_aC$, then $II_a$ is a diameter of $omega_a$.
Check that $M_aB=M_aC$. This is easy.
Prove that $A$, $I$, $M_a$, and $I_a$ are collinear. This is also easy. (Recall that $I$ and $I_a$ are on the internal angular bisector of $angle BAC$.)
Use (2) and (3) to verify that $M_aB=M_aI$. This implies that $M_a$ is the circumcenter of the triangle $IBC$. However, as $I$, $B$, $I_a$, and $C$ are concyclic, the claim follows.
answered Nov 13 at 22:57
Batominovski
31.3k23187
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
By angle chasing, you can prove the statement. Here are some steps to do so.
Let $ABC$ be a triangle with circumcircle $Gamma$. Write $I$ and $I_a$, respectively, for the incenter and the excenter opposite to the vertex $A$ of the triangle $ABC$. Denote by $M_a$ the midpoint of the arc $BC$ of $Gamma$ that does not contain $A$.
Show that $angle IBI_a=dfrac{pi}{2}=angle ICI_a$. As a consequence, $IBI_aC$ is a cyclic quadrilateral. If $omega_a$ is the circumcircle of the quadrilateral $IBI_aC$, then $II_a$ is a diameter of $omega_a$.
Check that $M_aB=M_aC$. This is easy.
Prove that $A$, $I$, $M_a$, and $I_a$ are collinear. This is also easy. (Recall that $I$ and $I_a$ are on the internal angular bisector of $angle BAC$.)
Use (2) and (3) to verify that $M_aB=M_aI$. This implies that $M_a$ is the circumcenter of the triangle $IBC$. However, as $I$, $B$, $I_a$, and $C$ are concyclic, the claim follows.
answered Nov 13 at 22:57
Batominovski
31.3k23187
add a comment |
up vote
1
down vote
By angle chasing, you can prove the statement. Here are some steps to do so.
Let $ABC$ be a triangle with circumcircle $Gamma$. Write $I$ and $I_a$, respectively, for the incenter and the excenter opposite to the vertex $A$ of the triangle $ABC$. Denote by $M_a$ the midpoint of the arc $BC$ of $Gamma$ that does not contain $A$.
Show that $angle IBI_a=dfrac{pi}{2}=angle ICI_a$. As a consequence, $IBI_aC$ is a cyclic quadrilateral. If $omega_a$ is the circumcircle of the quadrilateral $IBI_aC$, then $II_a$ is a diameter of $omega_a$.
Check that $M_aB=M_aC$. This is easy.
Prove that $A$, $I$, $M_a$, and $I_a$ are collinear. This is also easy. (Recall that $I$ and $I_a$ are on the internal angular bisector of $angle BAC$.)
Use (2) and (3) to verify that $M_aB=M_aI$. This implies that $M_a$ is the circumcenter of the triangle $IBC$. However, as $I$, $B$, $I_a$, and $C$ are concyclic, the claim follows.
answered Nov 13 at 22:57
Batominovski
31.3k23187
add a comment |
up vote
1
down vote
up vote
1
down vote
By angle chasing, you can prove the statement. Here are some steps to do so.
Let $ABC$ be a triangle with circumcircle $Gamma$. Write $I$ and $I_a$, respectively, for the incenter and the excenter opposite to the vertex $A$ of the triangle $ABC$. Denote by $M_a$ the midpoint of the arc $BC$ of $Gamma$ that does not contain $A$.
Show that $angle IBI_a=dfrac{pi}{2}=angle ICI_a$. As a consequence, $IBI_aC$ is a cyclic quadrilateral. If $omega_a$ is the circumcircle of the quadrilateral $IBI_aC$, then $II_a$ is a diameter of $omega_a$.
Check that $M_aB=M_aC$. This is easy.
Prove that $A$, $I$, $M_a$, and $I_a$ are collinear. This is also easy. (Recall that $I$ and $I_a$ are on the internal angular bisector of $angle BAC$.)
Use (2) and (3) to verify that $M_aB=M_aI$. This implies that $M_a$ is the circumcenter of the triangle $IBC$. However, as $I$, $B$, $I_a$, and $C$ are concyclic, the claim follows.
answered Nov 13 at 22:57
Batominovski
31.3k23187
By angle chasing, you can prove the statement. Here are some steps to do so.
Let $ABC$ be a triangle with circumcircle $Gamma$. Write $I$ and $I_a$, respectively, for the incenter and the excenter opposite to the vertex $A$ of the triangle $ABC$. Denote by $M_a$ the midpoint of the arc $BC$ of $Gamma$ that does not contain $A$.
Show that $angle IBI_a=dfrac{pi}{2}=angle ICI_a$. As a consequence, $IBI_aC$ is a cyclic quadrilateral. If $omega_a$ is the circumcircle of the quadrilateral $IBI_aC$, then $II_a$ is a diameter of $omega_a$.
Check that $M_aB=M_aC$. This is easy.
Prove that $A$, $I$, $M_a$, and $I_a$ are collinear. This is also easy. (Recall that $I$ and $I_a$ are on the internal angular bisector of $angle BAC$.)
Use (2) and (3) to verify that $M_aB=M_aI$. This implies that $M_a$ is the circumcenter of the triangle $IBC$. However, as $I$, $B$, $I_a$, and $C$ are concyclic, the claim follows.
answered Nov 13 at 22:57
Batominovski
31.3k23187
answered Nov 13 at 22:57
Batominovski
31.3k23187
answered Nov 13 at 22:57
Batominovski
31.3k23187
answered Nov 13 at 22:57
Batominovski
31.3k23187
31.3k23187
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997457%2fproove-line-circumscribed-circle-incircle-excircle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
