Mixing
Let S be a semicircle with diameter AB. The point C lies on the diameter AB and points E and D lie on the arc...
$begingroup$
Let $S$ be a semicircle with diameter $AB$. The point $C$ lies on the diameter $AB$ and points $E$ and $D$ lie on the arc $BA$, with $E$ between $B$ and $D$. Let the tangents to $S$ at $D$ and $E$ meets at $F$. Suppose that $angle ACD = angle ECB$. Prove that $angle EFD = angle ACD + angle ECB$.
geometry
edited Jan 18 at 19:21
Andrei
12.4k21128
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
$endgroup$
add a comment |
$begingroup$
Let $S$ be a semicircle with diameter $AB$. The point $C$ lies on the diameter $AB$ and points $E$ and $D$ lie on the arc $BA$, with $E$ between $B$ and $D$. Let the tangents to $S$ at $D$ and $E$ meets at $F$. Suppose that $angle ACD = angle ECB$. Prove that $angle EFD = angle ACD + angle ECB$.
geometry
edited Jan 18 at 19:21
Andrei
12.4k21128
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
$endgroup$
$begingroup$
This is algebra?
$endgroup$
– greedoid
Jan 18 at 19:20
$begingroup$
What have you tried? Are you stuck on some concept or calculation? People would like to help, not solve the problem for you.
$endgroup$
– Andrei
Jan 18 at 19:24
$begingroup$
I edited a bit my answer, hope it is more clear now.
$endgroup$
– GReyes
Jan 19 at 18:21
add a comment |
$begingroup$
Let $S$ be a semicircle with diameter $AB$. The point $C$ lies on the diameter $AB$ and points $E$ and $D$ lie on the arc $BA$, with $E$ between $B$ and $D$. Let the tangents to $S$ at $D$ and $E$ meets at $F$. Suppose that $angle ACD = angle ECB$. Prove that $angle EFD = angle ACD + angle ECB$.
geometry
edited Jan 18 at 19:21
Andrei
12.4k21128
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
$endgroup$
Let $S$ be a semicircle with diameter $AB$. The point $C$ lies on the diameter $AB$ and points $E$ and $D$ lie on the arc $BA$, with $E$ between $B$ and $D$. Let the tangents to $S$ at $D$ and $E$ meets at $F$. Suppose that $angle ACD = angle ECB$. Prove that $angle EFD = angle ACD + angle ECB$.
geometry
geometry
edited Jan 18 at 19:21
Andrei
12.4k21128
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
edited Jan 18 at 19:21
Andrei
12.4k21128
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
edited Jan 18 at 19:21
Andrei
12.4k21128
edited Jan 18 at 19:21
Andrei
12.4k21128
edited Jan 18 at 19:21
Andrei
12.4k21128
12.4k21128
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
asked Jan 18 at 19:12
Anson ChanAnson Chan
162
162
$begingroup$
This is algebra?
$endgroup$
– greedoid
Jan 18 at 19:20
$begingroup$
What have you tried? Are you stuck on some concept or calculation? People would like to help, not solve the problem for you.
$endgroup$
– Andrei
Jan 18 at 19:24
$begingroup$
I edited a bit my answer, hope it is more clear now.
$endgroup$
– GReyes
Jan 19 at 18:21
add a comment |
$begingroup$
This is algebra?
$endgroup$
– greedoid
Jan 18 at 19:20
$begingroup$
What have you tried? Are you stuck on some concept or calculation? People would like to help, not solve the problem for you.
$endgroup$
– Andrei
Jan 18 at 19:24
$begingroup$
I edited a bit my answer, hope it is more clear now.
$endgroup$
– GReyes
Jan 19 at 18:21
$begingroup$
This is algebra?
$endgroup$
– greedoid
Jan 18 at 19:20
$begingroup$
This is algebra?
$endgroup$
– greedoid
Jan 18 at 19:20
$begingroup$
What have you tried? Are you stuck on some concept or calculation? People would like to help, not solve the problem for you.
$endgroup$
– Andrei
Jan 18 at 19:24
$begingroup$
What have you tried? Are you stuck on some concept or calculation? People would like to help, not solve the problem for you.
$endgroup$
– Andrei
Jan 18 at 19:24
$begingroup$
I edited a bit my answer, hope it is more clear now.
$endgroup$
– GReyes
Jan 19 at 18:21
$begingroup$
I edited a bit my answer, hope it is more clear now.
$endgroup$
– GReyes
Jan 19 at 18:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will assume that $C$ is closer to $B$ than to $A$, the other case is symmetric. If you join $D$ and $E$ to the center $O$, the quadrilateral $OEFD$ is inscriptible and therefore $angle DOE=pi-angle DFE$. . You want to prove that the quadrilateral $DOCE$ is inscriptible as well (the circumscribed circle will be the same as before) and therefore $angle DCE=angle DOE$, since they are both inscribed on the same chord. But clearly $angle DCE=pi-angle ACD-angle ECB$ and your assertion follows.
To prove that the quadrilateral $DOCE$ is inscriptible you can just extend $DC$ to the other side of the circle (call $G$ the point of intersection) and observe that, by the given relation $angle ACD=angle ECB$ you have that $E$ and $G$ are symmetric with respect to the diameter $AB$. That symmetry implies that $angle OEC=angle CGO$. You also have that $angle CGO=angle ODC$ since the triangle $GOD$ is isosceles. This establishes that $angle OEC=angle ODC$ which proves that $OCED$ is inscriptible, hence the desired assertion.
Hope this helps.
answered Jan 19 at 6:55
GReyesGReyes
1,57015
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
I will assume that $C$ is closer to $B$ than to $A$, the other case is symmetric. If you join $D$ and $E$ to the center $O$, the quadrilateral $OEFD$ is inscriptible and therefore $angle DOE=pi-angle DFE$. . You want to prove that the quadrilateral $DOCE$ is inscriptible as well (the circumscribed circle will be the same as before) and therefore $angle DCE=angle DOE$, since they are both inscribed on the same chord. But clearly $angle DCE=pi-angle ACD-angle ECB$ and your assertion follows.
To prove that the quadrilateral $DOCE$ is inscriptible you can just extend $DC$ to the other side of the circle (call $G$ the point of intersection) and observe that, by the given relation $angle ACD=angle ECB$ you have that $E$ and $G$ are symmetric with respect to the diameter $AB$. That symmetry implies that $angle OEC=angle CGO$. You also have that $angle CGO=angle ODC$ since the triangle $GOD$ is isosceles. This establishes that $angle OEC=angle ODC$ which proves that $OCED$ is inscriptible, hence the desired assertion.
Hope this helps.
answered Jan 19 at 6:55
GReyesGReyes
1,57015
$endgroup$
add a comment |
$begingroup$
I will assume that $C$ is closer to $B$ than to $A$, the other case is symmetric. If you join $D$ and $E$ to the center $O$, the quadrilateral $OEFD$ is inscriptible and therefore $angle DOE=pi-angle DFE$. . You want to prove that the quadrilateral $DOCE$ is inscriptible as well (the circumscribed circle will be the same as before) and therefore $angle DCE=angle DOE$, since they are both inscribed on the same chord. But clearly $angle DCE=pi-angle ACD-angle ECB$ and your assertion follows.
To prove that the quadrilateral $DOCE$ is inscriptible you can just extend $DC$ to the other side of the circle (call $G$ the point of intersection) and observe that, by the given relation $angle ACD=angle ECB$ you have that $E$ and $G$ are symmetric with respect to the diameter $AB$. That symmetry implies that $angle OEC=angle CGO$. You also have that $angle CGO=angle ODC$ since the triangle $GOD$ is isosceles. This establishes that $angle OEC=angle ODC$ which proves that $OCED$ is inscriptible, hence the desired assertion.
Hope this helps.
answered Jan 19 at 6:55
GReyesGReyes
1,57015
$endgroup$
add a comment |
$begingroup$
I will assume that $C$ is closer to $B$ than to $A$, the other case is symmetric. If you join $D$ and $E$ to the center $O$, the quadrilateral $OEFD$ is inscriptible and therefore $angle DOE=pi-angle DFE$. . You want to prove that the quadrilateral $DOCE$ is inscriptible as well (the circumscribed circle will be the same as before) and therefore $angle DCE=angle DOE$, since they are both inscribed on the same chord. But clearly $angle DCE=pi-angle ACD-angle ECB$ and your assertion follows.
To prove that the quadrilateral $DOCE$ is inscriptible you can just extend $DC$ to the other side of the circle (call $G$ the point of intersection) and observe that, by the given relation $angle ACD=angle ECB$ you have that $E$ and $G$ are symmetric with respect to the diameter $AB$. That symmetry implies that $angle OEC=angle CGO$. You also have that $angle CGO=angle ODC$ since the triangle $GOD$ is isosceles. This establishes that $angle OEC=angle ODC$ which proves that $OCED$ is inscriptible, hence the desired assertion.
Hope this helps.
answered Jan 19 at 6:55
GReyesGReyes
1,57015
$endgroup$
I will assume that $C$ is closer to $B$ than to $A$, the other case is symmetric. If you join $D$ and $E$ to the center $O$, the quadrilateral $OEFD$ is inscriptible and therefore $angle DOE=pi-angle DFE$. . You want to prove that the quadrilateral $DOCE$ is inscriptible as well (the circumscribed circle will be the same as before) and therefore $angle DCE=angle DOE$, since they are both inscribed on the same chord. But clearly $angle DCE=pi-angle ACD-angle ECB$ and your assertion follows.
To prove that the quadrilateral $DOCE$ is inscriptible you can just extend $DC$ to the other side of the circle (call $G$ the point of intersection) and observe that, by the given relation $angle ACD=angle ECB$ you have that $E$ and $G$ are symmetric with respect to the diameter $AB$. That symmetry implies that $angle OEC=angle CGO$. You also have that $angle CGO=angle ODC$ since the triangle $GOD$ is isosceles. This establishes that $angle OEC=angle ODC$ which proves that $OCED$ is inscriptible, hence the desired assertion.
Hope this helps.
answered Jan 19 at 6:55
GReyesGReyes
1,57015
edited Jan 19 at 18:19
answered Jan 19 at 6:55
GReyesGReyes
1,57015
answered Jan 19 at 6:55
GReyesGReyes
1,57015
answered Jan 19 at 6:55
GReyesGReyes
1,57015
1,57015
add a comment |
add a comment |
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$begingroup$
This is algebra?
$endgroup$
– greedoid
Jan 18 at 19:20
$begingroup$
What have you tried? Are you stuck on some concept or calculation? People would like to help, not solve the problem for you.
$endgroup$
– Andrei
Jan 18 at 19:24
$begingroup$
I edited a bit my answer, hope it is more clear now.
$endgroup$
– GReyes
Jan 19 at 18:21