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Circle Geometry Question Involving Two Tangents and a Set of Parallel Lines
Clash Royale CLAN TAG#URR8PPP
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Tangents $OA$ and $OB$ are drawn to a circle from an external point $O$. Through the point $A$, a chord $AC$ is drawn parallel to the tangent $OB$ and $OC$ passes through the circle at $E$.
I am required to show that A$F$ bisects $OB$.
My approach thus far has been to observe that triangles $AFO$ and $OFE$ are similar, and also $AEB$ is similar to $BEO$.
However I am not able to reach an appropriate proof... Please help!
geometry euclidean-geometry circle
edited Jul 22 at 9:59
greedoid
26.2k93473
asked Jul 22 at 9:43
Hugh Entwistle
614115
add a comment |Â
up vote
1
down vote
favorite
Tangents $OA$ and $OB$ are drawn to a circle from an external point $O$. Through the point $A$, a chord $AC$ is drawn parallel to the tangent $OB$ and $OC$ passes through the circle at $E$.
I am required to show that A$F$ bisects $OB$.
My approach thus far has been to observe that triangles $AFO$ and $OFE$ are similar, and also $AEB$ is similar to $BEO$.
However I am not able to reach an appropriate proof... Please help!
geometry euclidean-geometry circle
edited Jul 22 at 9:59
greedoid
26.2k93473
asked Jul 22 at 9:43
Hugh Entwistle
614115
I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO.
– Hugh Entwistle
Jul 22 at 9:48
Math.SE demands you to show at least some work on your question. What have you tried so far?
– Cargobob
Jul 22 at 9:50
1
I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire.
– Hugh Entwistle
Jul 22 at 9:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Tangents $OA$ and $OB$ are drawn to a circle from an external point $O$. Through the point $A$, a chord $AC$ is drawn parallel to the tangent $OB$ and $OC$ passes through the circle at $E$.
I am required to show that A$F$ bisects $OB$.
My approach thus far has been to observe that triangles $AFO$ and $OFE$ are similar, and also $AEB$ is similar to $BEO$.
However I am not able to reach an appropriate proof... Please help!
geometry euclidean-geometry circle
edited Jul 22 at 9:59
greedoid
26.2k93473
asked Jul 22 at 9:43
Hugh Entwistle
614115
Tangents $OA$ and $OB$ are drawn to a circle from an external point $O$. Through the point $A$, a chord $AC$ is drawn parallel to the tangent $OB$ and $OC$ passes through the circle at $E$.
I am required to show that A$F$ bisects $OB$.
My approach thus far has been to observe that triangles $AFO$ and $OFE$ are similar, and also $AEB$ is similar to $BEO$.
However I am not able to reach an appropriate proof... Please help!
geometry euclidean-geometry circle
edited Jul 22 at 9:59
greedoid
26.2k93473
asked Jul 22 at 9:43
Hugh Entwistle
614115
edited Jul 22 at 9:59
greedoid
26.2k93473
edited Jul 22 at 9:59
greedoid
26.2k93473
edited Jul 22 at 9:59
greedoid
26.2k93473
26.2k93473
asked Jul 22 at 9:43
Hugh Entwistle
614115
asked Jul 22 at 9:43
Hugh Entwistle
614115
asked Jul 22 at 9:43
Hugh Entwistle
614115
614115
I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO.
– Hugh Entwistle
Jul 22 at 9:48
Math.SE demands you to show at least some work on your question. What have you tried so far?
– Cargobob
Jul 22 at 9:50
1
I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire.
– Hugh Entwistle
Jul 22 at 9:52
add a comment |Â
I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO.
– Hugh Entwistle
Jul 22 at 9:48
Math.SE demands you to show at least some work on your question. What have you tried so far?
– Cargobob
Jul 22 at 9:50
1
I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire.
– Hugh Entwistle
Jul 22 at 9:52
I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO.
– Hugh Entwistle
Jul 22 at 9:48
I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO.
– Hugh Entwistle
Jul 22 at 9:48
Math.SE demands you to show at least some work on your question. What have you tried so far?
– Cargobob
Jul 22 at 9:50
Math.SE demands you to show at least some work on your question. What have you tried so far?
– Cargobob
Jul 22 at 9:50
1
1
I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire.
– Hugh Entwistle
Jul 22 at 9:52
I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire.
– Hugh Entwistle
Jul 22 at 9:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Since $angle EAO = angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FEcdot FA$$
and by PoP on a given circle we have $$FB^2 = FEcdot FA$$
so $FB = FO$.
answered Jul 22 at 9:54
greedoid
26.2k93473
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Since $angle EAO = angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FEcdot FA$$
and by PoP on a given circle we have $$FB^2 = FEcdot FA$$
so $FB = FO$.
answered Jul 22 at 9:54
greedoid
26.2k93473
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
add a comment |Â
up vote
0
down vote
accepted
Since $angle EAO = angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FEcdot FA$$
and by PoP on a given circle we have $$FB^2 = FEcdot FA$$
so $FB = FO$.
answered Jul 22 at 9:54
greedoid
26.2k93473
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since $angle EAO = angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FEcdot FA$$
and by PoP on a given circle we have $$FB^2 = FEcdot FA$$
so $FB = FO$.
answered Jul 22 at 9:54
greedoid
26.2k93473
Since $angle EAO = angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FEcdot FA$$
and by PoP on a given circle we have $$FB^2 = FEcdot FA$$
so $FB = FO$.
answered Jul 22 at 9:54
greedoid
26.2k93473
answered Jul 22 at 9:54
greedoid
26.2k93473
answered Jul 22 at 9:54
greedoid
26.2k93473
answered Jul 22 at 9:54
greedoid
26.2k93473
26.2k93473
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
add a comment |Â
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation).
– Jean Marie
Jul 22 at 18:01
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
Yes, a saw it elsewhere
– greedoid
Jul 22 at 18:02
add a comment |Â
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I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO.
– Hugh Entwistle
Jul 22 at 9:48
Math.SE demands you to show at least some work on your question. What have you tried so far?
– Cargobob
Jul 22 at 9:50
1
I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire.
– Hugh Entwistle
Jul 22 at 9:52