Circumscribed triangle and tangent
$begingroup$
In the given diagram , B C T are on circle and AT tangent to the circle at T , if AB =3 and BC=4 find $${AB +ATover AT+AC}$$
](https://i.stack.imgur.com/Vw9AA.jpg)
geometry euclidean-geometry
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
asked Dec 26 '18 at 19:30
RandinRandin
347116
$endgroup$
add a comment |
$begingroup$
In the given diagram , B C T are on circle and AT tangent to the circle at T , if AB =3 and BC=4 find $${AB +ATover AT+AC}$$
geometry euclidean-geometry
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
asked Dec 26 '18 at 19:30
RandinRandin
347116
$endgroup$
$begingroup$
This remark has no effect on the answer to the question, but terminology matters. In the diagram I see an inscribed triangle. I see nothing I would call a circumscribed triangle.
$endgroup$
– David K
Dec 27 '18 at 20:27
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
$endgroup$
– Randin
Dec 28 '18 at 0:03
$begingroup$
That’s a good observation about the similar triangles. I would just call them “triangles” or “similar triangles” to avoid confusion, however.
$endgroup$
– David K
Dec 28 '18 at 0:06
add a comment |
$begingroup$
In the given diagram , B C T are on circle and AT tangent to the circle at T , if AB =3 and BC=4 find $${AB +ATover AT+AC}$$
geometry euclidean-geometry
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
asked Dec 26 '18 at 19:30
RandinRandin
347116
$endgroup$
In the given diagram , B C T are on circle and AT tangent to the circle at T , if AB =3 and BC=4 find $${AB +ATover AT+AC}$$
geometry euclidean-geometry
geometry euclidean-geometry
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
asked Dec 26 '18 at 19:30
RandinRandin
347116
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
asked Dec 26 '18 at 19:30
RandinRandin
347116
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
edited Dec 29 '18 at 10:48
Maria Mazur
47.1k1260120
47.1k1260120
asked Dec 26 '18 at 19:30
RandinRandin
347116
asked Dec 26 '18 at 19:30
RandinRandin
347116
asked Dec 26 '18 at 19:30
RandinRandin
347116
347116
$begingroup$
This remark has no effect on the answer to the question, but terminology matters. In the diagram I see an inscribed triangle. I see nothing I would call a circumscribed triangle.
$endgroup$
– David K
Dec 27 '18 at 20:27
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
$endgroup$
– Randin
Dec 28 '18 at 0:03
$begingroup$
That’s a good observation about the similar triangles. I would just call them “triangles” or “similar triangles” to avoid confusion, however.
$endgroup$
– David K
Dec 28 '18 at 0:06
add a comment |
$begingroup$
This remark has no effect on the answer to the question, but terminology matters. In the diagram I see an inscribed triangle. I see nothing I would call a circumscribed triangle.
$endgroup$
– David K
Dec 27 '18 at 20:27
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
$endgroup$
– Randin
Dec 28 '18 at 0:03
$begingroup$
That’s a good observation about the similar triangles. I would just call them “triangles” or “similar triangles” to avoid confusion, however.
$endgroup$
– David K
Dec 28 '18 at 0:06
$begingroup$
This remark has no effect on the answer to the question, but terminology matters. In the diagram I see an inscribed triangle. I see nothing I would call a circumscribed triangle.
$endgroup$
– David K
Dec 27 '18 at 20:27
$begingroup$
This remark has no effect on the answer to the question, but terminology matters. In the diagram I see an inscribed triangle. I see nothing I would call a circumscribed triangle.
$endgroup$
– David K
Dec 27 '18 at 20:27
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
$endgroup$
– Randin
Dec 28 '18 at 0:03
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
$endgroup$
– Randin
Dec 28 '18 at 0:03
$begingroup$
That’s a good observation about the similar triangles. I would just call them “triangles” or “similar triangles” to avoid confusion, however.
$endgroup$
– David K
Dec 28 '18 at 0:06
$begingroup$
That’s a good observation about the similar triangles. I would just call them “triangles” or “similar triangles” to avoid confusion, however.
$endgroup$
– David K
Dec 28 '18 at 0:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you know the power of a point, then we have (with respect to point $A$ and a given circle): $$AT^2=ABcdot AC = 3cdot 7 =21$$
so $${AB+ATover AC+AT}=...$$
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
$endgroup$
1
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
add a comment |
$begingroup$
Angles $ACT$ and $BTA$ both measure half the intercepted arc $TB$. So the triangles vlcontaining these angles, both also containing angle $TCA$, are similar. From the proportionality of corresponding sides of similar triangles, conclude that $|AT|$ is the geometric mean of $|AB|$ and $|AC|$. Render $|AC|=7$ from betweenness and the rest is just substituting the numbers.
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
add a comment |
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
answered Dec 28 '18 at 0:04
RandinRandin
347116
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you know the power of a point, then we have (with respect to point $A$ and a given circle): $$AT^2=ABcdot AC = 3cdot 7 =21$$
so $${AB+ATover AC+AT}=...$$
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
$endgroup$
1
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
add a comment |
$begingroup$
If you know the power of a point, then we have (with respect to point $A$ and a given circle): $$AT^2=ABcdot AC = 3cdot 7 =21$$
so $${AB+ATover AC+AT}=...$$
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
$endgroup$
1
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
add a comment |
$begingroup$
If you know the power of a point, then we have (with respect to point $A$ and a given circle): $$AT^2=ABcdot AC = 3cdot 7 =21$$
so $${AB+ATover AC+AT}=...$$
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
$endgroup$
If you know the power of a point, then we have (with respect to point $A$ and a given circle): $$AT^2=ABcdot AC = 3cdot 7 =21$$
so $${AB+ATover AC+AT}=...$$
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
answered Dec 26 '18 at 20:53
Maria MazurMaria Mazur
47.1k1260120
47.1k1260120
1
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
add a comment |
1
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
1
1
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
$begingroup$
YaY I actually derived that ! from similar triangles on the two triangles right ?
$endgroup$
– Randin
Dec 26 '18 at 21:19
add a comment |
$begingroup$
Angles $ACT$ and $BTA$ both measure half the intercepted arc $TB$. So the triangles vlcontaining these angles, both also containing angle $TCA$, are similar. From the proportionality of corresponding sides of similar triangles, conclude that $|AT|$ is the geometric mean of $|AB|$ and $|AC|$. Render $|AC|=7$ from betweenness and the rest is just substituting the numbers.
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
add a comment |
$begingroup$
Angles $ACT$ and $BTA$ both measure half the intercepted arc $TB$. So the triangles vlcontaining these angles, both also containing angle $TCA$, are similar. From the proportionality of corresponding sides of similar triangles, conclude that $|AT|$ is the geometric mean of $|AB|$ and $|AC|$. Render $|AC|=7$ from betweenness and the rest is just substituting the numbers.
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
add a comment |
$begingroup$
Angles $ACT$ and $BTA$ both measure half the intercepted arc $TB$. So the triangles vlcontaining these angles, both also containing angle $TCA$, are similar. From the proportionality of corresponding sides of similar triangles, conclude that $|AT|$ is the geometric mean of $|AB|$ and $|AC|$. Render $|AC|=7$ from betweenness and the rest is just substituting the numbers.
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
Angles $ACT$ and $BTA$ both measure half the intercepted arc $TB$. So the triangles vlcontaining these angles, both also containing angle $TCA$, are similar. From the proportionality of corresponding sides of similar triangles, conclude that $|AT|$ is the geometric mean of $|AB|$ and $|AC|$. Render $|AC|=7$ from betweenness and the rest is just substituting the numbers.
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
answered Dec 26 '18 at 19:55
Oscar LanziOscar Lanzi
13.2k12136
13.2k12136
add a comment |
add a comment |
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
answered Dec 28 '18 at 0:04
RandinRandin
347116
$endgroup$
add a comment |
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
answered Dec 28 '18 at 0:04
RandinRandin
347116
$endgroup$
add a comment |
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
answered Dec 28 '18 at 0:04
RandinRandin
347116
$endgroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
answered Dec 28 '18 at 0:04
RandinRandin
347116
answered Dec 28 '18 at 0:04
RandinRandin
347116
answered Dec 28 '18 at 0:04
RandinRandin
347116
answered Dec 28 '18 at 0:04
RandinRandin
347116
347116
add a comment |
add a comment |
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$begingroup$
This remark has no effect on the answer to the question, but terminology matters. In the diagram I see an inscribed triangle. I see nothing I would call a circumscribed triangle.
$endgroup$
– David K
Dec 27 '18 at 20:27
$begingroup$
My glitch David K, the key thing i saw that turns out to be ' power of a point' was similar triangles ATB & ATC ( along with THC ) gives side AT/ AB=AC/AT
$endgroup$
– Randin
Dec 28 '18 at 0:03
$begingroup$
That’s a good observation about the similar triangles. I would just call them “triangles” or “similar triangles” to avoid confusion, however.
$endgroup$
– David K
Dec 28 '18 at 0:06