Mixing
Angle at circumference and centre relationship
$begingroup$
An angle made at the circumference of a circle is half of the angle made at the centre.
But is the converse also true? i.e. suppose $triangle ABC$ is a triangle and angle $angle ABC=45^circ$. Let $K$ be a point inside triangle $triangle ABC$ and $angle AKC= 90^circ$. Then can we say that $K$ is circumcentre of $triangle ABC$?
geometry
edited Jan 30 at 19:14
Infiaria
47011
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
$endgroup$
add a comment |
$begingroup$
An angle made at the circumference of a circle is half of the angle made at the centre.
But is the converse also true? i.e. suppose $triangle ABC$ is a triangle and angle $angle ABC=45^circ$. Let $K$ be a point inside triangle $triangle ABC$ and $angle AKC= 90^circ$. Then can we say that $K$ is circumcentre of $triangle ABC$?
geometry
edited Jan 30 at 19:14
Infiaria
47011
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
$endgroup$
1
$begingroup$
No, there are infinitely many right triangles that have hypotenuse equal to $AC$ but only one circumcentre..
$endgroup$
– Vasya
Jan 30 at 18:20
add a comment |
$begingroup$
An angle made at the circumference of a circle is half of the angle made at the centre.
But is the converse also true? i.e. suppose $triangle ABC$ is a triangle and angle $angle ABC=45^circ$. Let $K$ be a point inside triangle $triangle ABC$ and $angle AKC= 90^circ$. Then can we say that $K$ is circumcentre of $triangle ABC$?
geometry
edited Jan 30 at 19:14
Infiaria
47011
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
$endgroup$
An angle made at the circumference of a circle is half of the angle made at the centre.
But is the converse also true? i.e. suppose $triangle ABC$ is a triangle and angle $angle ABC=45^circ$. Let $K$ be a point inside triangle $triangle ABC$ and $angle AKC= 90^circ$. Then can we say that $K$ is circumcentre of $triangle ABC$?
geometry
geometry
edited Jan 30 at 19:14
Infiaria
47011
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
edited Jan 30 at 19:14
Infiaria
47011
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
edited Jan 30 at 19:14
Infiaria
47011
edited Jan 30 at 19:14
Infiaria
47011
edited Jan 30 at 19:14
Infiaria
47011
47011
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
asked Jan 30 at 18:04
Kshitij SinghKshitij Singh
1326
1326
1
$begingroup$
No, there are infinitely many right triangles that have hypotenuse equal to $AC$ but only one circumcentre..
$endgroup$
– Vasya
Jan 30 at 18:20
add a comment |
1
$begingroup$
No, there are infinitely many right triangles that have hypotenuse equal to $AC$ but only one circumcentre..
$endgroup$
– Vasya
Jan 30 at 18:20
1
1
$begingroup$
No, there are infinitely many right triangles that have hypotenuse equal to $AC$ but only one circumcentre..
$endgroup$
– Vasya
Jan 30 at 18:20
$begingroup$
No, there are infinitely many right triangles that have hypotenuse equal to $AC$ but only one circumcentre..
$endgroup$
– Vasya
Jan 30 at 18:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No.
If we define $O$ as the circumcenter of $ΔABC$, then $angle AOC=frac{pi}{2}$.
Hence (by angles on the same arc), any point $K$ on $(AOC)$ suffices,
as $angle AKB = angle AOB = frac{pi}{2}$.
However, it is true that a point $K$ is the circumcentre of $Delta ABC$ iff
$angle AKB = angle AOB$,
$angle BKC = angle BOC$ and
$angle CKA = angle COA$.
(As then, $K$ must be on $(AOB)$, $(BOC)$ and $(COA)$.)
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
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add a comment |
$begingroup$
The converse of "angle at center = twice angle at circumference" is not necessarily true. Figure 1 shows a counter-example.
Figure 2 shows there is such a possibility provided (1) O is the center of the target circle with OA = OB, its radius; and (2) Both O and $P_1$ must lie on the same side of the chord AB.
Violating (2) will possibly end-up with the point $P_2$ shown in figure 3.
answered Jan 31 at 3:45
MickMick
12k31641
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
No.
If we define $O$ as the circumcenter of $ΔABC$, then $angle AOC=frac{pi}{2}$.
Hence (by angles on the same arc), any point $K$ on $(AOC)$ suffices,
as $angle AKB = angle AOB = frac{pi}{2}$.
However, it is true that a point $K$ is the circumcentre of $Delta ABC$ iff
$angle AKB = angle AOB$,
$angle BKC = angle BOC$ and
$angle CKA = angle COA$.
(As then, $K$ must be on $(AOB)$, $(BOC)$ and $(COA)$.)
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
add a comment |
$begingroup$
No.
If we define $O$ as the circumcenter of $ΔABC$, then $angle AOC=frac{pi}{2}$.
Hence (by angles on the same arc), any point $K$ on $(AOC)$ suffices,
as $angle AKB = angle AOB = frac{pi}{2}$.
However, it is true that a point $K$ is the circumcentre of $Delta ABC$ iff
$angle AKB = angle AOB$,
$angle BKC = angle BOC$ and
$angle CKA = angle COA$.
(As then, $K$ must be on $(AOB)$, $(BOC)$ and $(COA)$.)
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
add a comment |
$begingroup$
No.
If we define $O$ as the circumcenter of $ΔABC$, then $angle AOC=frac{pi}{2}$.
Hence (by angles on the same arc), any point $K$ on $(AOC)$ suffices,
as $angle AKB = angle AOB = frac{pi}{2}$.
However, it is true that a point $K$ is the circumcentre of $Delta ABC$ iff
$angle AKB = angle AOB$,
$angle BKC = angle BOC$ and
$angle CKA = angle COA$.
(As then, $K$ must be on $(AOB)$, $(BOC)$ and $(COA)$.)
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
$endgroup$
No.
If we define $O$ as the circumcenter of $ΔABC$, then $angle AOC=frac{pi}{2}$.
Hence (by angles on the same arc), any point $K$ on $(AOC)$ suffices,
as $angle AKB = angle AOB = frac{pi}{2}$.
However, it is true that a point $K$ is the circumcentre of $Delta ABC$ iff
$angle AKB = angle AOB$,
$angle BKC = angle BOC$ and
$angle CKA = angle COA$.
(As then, $K$ must be on $(AOB)$, $(BOC)$ and $(COA)$.)
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
edited Jan 30 at 18:28
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
answered Jan 30 at 18:22
Jonas De SchouwerJonas De Schouwer
3769
3769
add a comment |
add a comment |
$begingroup$
The converse of "angle at center = twice angle at circumference" is not necessarily true. Figure 1 shows a counter-example.
Figure 2 shows there is such a possibility provided (1) O is the center of the target circle with OA = OB, its radius; and (2) Both O and $P_1$ must lie on the same side of the chord AB.
Violating (2) will possibly end-up with the point $P_2$ shown in figure 3.
answered Jan 31 at 3:45
MickMick
12k31641
$endgroup$
add a comment |
$begingroup$
The converse of "angle at center = twice angle at circumference" is not necessarily true. Figure 1 shows a counter-example.
Figure 2 shows there is such a possibility provided (1) O is the center of the target circle with OA = OB, its radius; and (2) Both O and $P_1$ must lie on the same side of the chord AB.
Violating (2) will possibly end-up with the point $P_2$ shown in figure 3.
answered Jan 31 at 3:45
MickMick
12k31641
$endgroup$
add a comment |
$begingroup$
The converse of "angle at center = twice angle at circumference" is not necessarily true. Figure 1 shows a counter-example.
Figure 2 shows there is such a possibility provided (1) O is the center of the target circle with OA = OB, its radius; and (2) Both O and $P_1$ must lie on the same side of the chord AB.
Violating (2) will possibly end-up with the point $P_2$ shown in figure 3.
answered Jan 31 at 3:45
MickMick
12k31641
$endgroup$
The converse of "angle at center = twice angle at circumference" is not necessarily true. Figure 1 shows a counter-example.
Figure 2 shows there is such a possibility provided (1) O is the center of the target circle with OA = OB, its radius; and (2) Both O and $P_1$ must lie on the same side of the chord AB.
Violating (2) will possibly end-up with the point $P_2$ shown in figure 3.
answered Jan 31 at 3:45
MickMick
12k31641
answered Jan 31 at 3:45
MickMick
12k31641
answered Jan 31 at 3:45
MickMick
12k31641
answered Jan 31 at 3:45
MickMick
12k31641
12k31641
add a comment |
add a comment |
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$begingroup$
No, there are infinitely many right triangles that have hypotenuse equal to $AC$ but only one circumcentre..
$endgroup$
– Vasya
Jan 30 at 18:20