Mixing
if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the...
$begingroup$
if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the circumcircle of triangle ABC?
D is the circumcenter of BIC
geometry euclidean-geometry triangle circle
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
asked Jan 14 at 23:57
user604720user604720
737
$endgroup$
add a comment |
$begingroup$
if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the circumcircle of triangle ABC?
D is the circumcenter of BIC
geometry euclidean-geometry triangle circle
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
asked Jan 14 at 23:57
user604720user604720
737
$endgroup$
$begingroup$
Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C.
$endgroup$
– user10354138
Jan 15 at 0:27
add a comment |
$begingroup$
if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the circumcircle of triangle ABC?
D is the circumcenter of BIC
geometry euclidean-geometry triangle circle
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
asked Jan 14 at 23:57
user604720user604720
737
$endgroup$
if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the circumcircle of triangle ABC?
D is the circumcenter of BIC
geometry euclidean-geometry triangle circle
geometry euclidean-geometry triangle circle
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
asked Jan 14 at 23:57
user604720user604720
737
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
asked Jan 14 at 23:57
user604720user604720
737
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
edited Jan 15 at 6:54
Michael Rozenberg
104k1891196
104k1891196
asked Jan 14 at 23:57
user604720user604720
737
asked Jan 14 at 23:57
user604720user604720
737
asked Jan 14 at 23:57
user604720user604720
737
737
$begingroup$
Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C.
$endgroup$
– user10354138
Jan 15 at 0:27
add a comment |
$begingroup$
Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C.
$endgroup$
– user10354138
Jan 15 at 0:27
$begingroup$
Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C.
$endgroup$
– user10354138
Jan 15 at 0:27
$begingroup$
Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C.
$endgroup$
– user10354138
Jan 15 at 0:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Join $AI$ and extend till it intersects circumcircle.
If possible Ray$AI$ does not intersect $D$ but at $D'$ .
Join $D'B$ and $D'C$ .
By angle chasing,
$angle BID' = angle IBD'$
And
$angle CID' = angle ICD'$.
Therefore, $BD'$ = $ID'$ = $CD'$.
$implies$ $D'$ is circumcentre of $triangle BIC$ .
As given $D$ is circumcentre of $triangle BIC$ .
Contradiction to considerations,
$D$ coincides $D'$.
Hence, proved.
answered Jan 17 at 13:19
RB MCPERB MCPE
262
$endgroup$
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
add a comment |
$begingroup$
In the standard notation $$measuredangle BIC=180^{circ}-frac{beta}{2}-frac{gamma}{2}=180^{circ}-frac{180^{circ}-alpha}{2}=90^{circ}+frac{alpha}{2},$$ which says that in the circumcircle of $Delta BIC$ the arc $BC$ is equal to $180^{circ}+alpha$.
Thus the arc $BIC$ is $$360^{circ}-left(180^{circ}+alpharight)=180^{circ}-alpha$$
and since $$measuredangle BDC+measuredangle BAC=180^{circ},$$we obtain that the circumcenter of $Delta BIC$ is placed on the circumcircle of $Delta ABC.$
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Join $AI$ and extend till it intersects circumcircle.
If possible Ray$AI$ does not intersect $D$ but at $D'$ .
Join $D'B$ and $D'C$ .
By angle chasing,
$angle BID' = angle IBD'$
And
$angle CID' = angle ICD'$.
Therefore, $BD'$ = $ID'$ = $CD'$.
$implies$ $D'$ is circumcentre of $triangle BIC$ .
As given $D$ is circumcentre of $triangle BIC$ .
Contradiction to considerations,
$D$ coincides $D'$.
Hence, proved.
answered Jan 17 at 13:19
RB MCPERB MCPE
262
$endgroup$
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
add a comment |
$begingroup$
Join $AI$ and extend till it intersects circumcircle.
If possible Ray$AI$ does not intersect $D$ but at $D'$ .
Join $D'B$ and $D'C$ .
By angle chasing,
$angle BID' = angle IBD'$
And
$angle CID' = angle ICD'$.
Therefore, $BD'$ = $ID'$ = $CD'$.
$implies$ $D'$ is circumcentre of $triangle BIC$ .
As given $D$ is circumcentre of $triangle BIC$ .
Contradiction to considerations,
$D$ coincides $D'$.
Hence, proved.
answered Jan 17 at 13:19
RB MCPERB MCPE
262
$endgroup$
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
add a comment |
$begingroup$
Join $AI$ and extend till it intersects circumcircle.
If possible Ray$AI$ does not intersect $D$ but at $D'$ .
Join $D'B$ and $D'C$ .
By angle chasing,
$angle BID' = angle IBD'$
And
$angle CID' = angle ICD'$.
Therefore, $BD'$ = $ID'$ = $CD'$.
$implies$ $D'$ is circumcentre of $triangle BIC$ .
As given $D$ is circumcentre of $triangle BIC$ .
Contradiction to considerations,
$D$ coincides $D'$.
Hence, proved.
answered Jan 17 at 13:19
RB MCPERB MCPE
262
$endgroup$
Join $AI$ and extend till it intersects circumcircle.
If possible Ray$AI$ does not intersect $D$ but at $D'$ .
Join $D'B$ and $D'C$ .
By angle chasing,
$angle BID' = angle IBD'$
And
$angle CID' = angle ICD'$.
Therefore, $BD'$ = $ID'$ = $CD'$.
$implies$ $D'$ is circumcentre of $triangle BIC$ .
As given $D$ is circumcentre of $triangle BIC$ .
Contradiction to considerations,
$D$ coincides $D'$.
Hence, proved.
answered Jan 17 at 13:19
RB MCPERB MCPE
262
answered Jan 17 at 13:19
RB MCPERB MCPE
262
answered Jan 17 at 13:19
RB MCPERB MCPE
262
answered Jan 17 at 13:19
RB MCPERB MCPE
262
262
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
add a comment |
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD'
$endgroup$
– user604720
Jan 20 at 19:13
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
$begingroup$
Let $angle AIB=x$and $angle ABI=y$. $angle BID'=x+y$ . $angle CBI=y$ and $angle IBD'=x $ as $square ABCD'$ is cyclic. Thus $angle BID'=angle IBD'= x+y$. Repeat the process and get $angle CID'=angle ICD'$.
$endgroup$
– RB MCPE
Jan 22 at 10:00
add a comment |
$begingroup$
In the standard notation $$measuredangle BIC=180^{circ}-frac{beta}{2}-frac{gamma}{2}=180^{circ}-frac{180^{circ}-alpha}{2}=90^{circ}+frac{alpha}{2},$$ which says that in the circumcircle of $Delta BIC$ the arc $BC$ is equal to $180^{circ}+alpha$.
Thus the arc $BIC$ is $$360^{circ}-left(180^{circ}+alpharight)=180^{circ}-alpha$$
and since $$measuredangle BDC+measuredangle BAC=180^{circ},$$we obtain that the circumcenter of $Delta BIC$ is placed on the circumcircle of $Delta ABC.$
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
add a comment |
$begingroup$
In the standard notation $$measuredangle BIC=180^{circ}-frac{beta}{2}-frac{gamma}{2}=180^{circ}-frac{180^{circ}-alpha}{2}=90^{circ}+frac{alpha}{2},$$ which says that in the circumcircle of $Delta BIC$ the arc $BC$ is equal to $180^{circ}+alpha$.
Thus the arc $BIC$ is $$360^{circ}-left(180^{circ}+alpharight)=180^{circ}-alpha$$
and since $$measuredangle BDC+measuredangle BAC=180^{circ},$$we obtain that the circumcenter of $Delta BIC$ is placed on the circumcircle of $Delta ABC.$
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
add a comment |
$begingroup$
In the standard notation $$measuredangle BIC=180^{circ}-frac{beta}{2}-frac{gamma}{2}=180^{circ}-frac{180^{circ}-alpha}{2}=90^{circ}+frac{alpha}{2},$$ which says that in the circumcircle of $Delta BIC$ the arc $BC$ is equal to $180^{circ}+alpha$.
Thus the arc $BIC$ is $$360^{circ}-left(180^{circ}+alpharight)=180^{circ}-alpha$$
and since $$measuredangle BDC+measuredangle BAC=180^{circ},$$we obtain that the circumcenter of $Delta BIC$ is placed on the circumcircle of $Delta ABC.$
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
$endgroup$
In the standard notation $$measuredangle BIC=180^{circ}-frac{beta}{2}-frac{gamma}{2}=180^{circ}-frac{180^{circ}-alpha}{2}=90^{circ}+frac{alpha}{2},$$ which says that in the circumcircle of $Delta BIC$ the arc $BC$ is equal to $180^{circ}+alpha$.
Thus the arc $BIC$ is $$360^{circ}-left(180^{circ}+alpharight)=180^{circ}-alpha$$
and since $$measuredangle BDC+measuredangle BAC=180^{circ},$$we obtain that the circumcenter of $Delta BIC$ is placed on the circumcircle of $Delta ABC.$
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
answered Jan 15 at 6:48
Michael RozenbergMichael Rozenberg
104k1891196
104k1891196
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
add a comment |
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
hi, quick question, what do β,γ, and α represent?
$endgroup$
– user604720
Jan 16 at 1:01
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
$begingroup$
@user604720 They are angles of the triangle. $measuredangle BAC=alpha$, $measuredangle ABC=beta$ and $measuredangle ACB=gamma.$
$endgroup$
– Michael Rozenberg
Jan 16 at 6:00
add a comment |
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$begingroup$
Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C.
$endgroup$
– user10354138
Jan 15 at 0:27