Proving a quadrilateral is cyclic
I am given that, for $JG$ an exterior angle bisector of $angle CGF$ parallel to the angle bisector of $angle FHE$, prove that $CDEF$ is a cyclic quadrilateral. I can prove that if the quadrilateral is cyclic, then the angle bisectors are parallel, but the reverse direction is proving troublesome...
geometry quadrilateral
asked Dec 10 '18 at 4:19
Derek Adams
515413
add a comment |
I am given that, for $JG$ an exterior angle bisector of $angle CGF$ parallel to the angle bisector of $angle FHE$, prove that $CDEF$ is a cyclic quadrilateral. I can prove that if the quadrilateral is cyclic, then the angle bisectors are parallel, but the reverse direction is proving troublesome...
geometry quadrilateral
asked Dec 10 '18 at 4:19
Derek Adams
515413
How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information
– user10354138
Dec 10 '18 at 4:27
$G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle.
– Derek Adams
Dec 10 '18 at 5:52
@DerekAdams, did you understand my answer?
– Anubhab Ghosal
Dec 10 '18 at 13:00
add a comment |
I am given that, for $JG$ an exterior angle bisector of $angle CGF$ parallel to the angle bisector of $angle FHE$, prove that $CDEF$ is a cyclic quadrilateral. I can prove that if the quadrilateral is cyclic, then the angle bisectors are parallel, but the reverse direction is proving troublesome...
geometry quadrilateral
asked Dec 10 '18 at 4:19
Derek Adams
515413
I am given that, for $JG$ an exterior angle bisector of $angle CGF$ parallel to the angle bisector of $angle FHE$, prove that $CDEF$ is a cyclic quadrilateral. I can prove that if the quadrilateral is cyclic, then the angle bisectors are parallel, but the reverse direction is proving troublesome...
geometry quadrilateral
geometry quadrilateral
asked Dec 10 '18 at 4:19
Derek Adams
515413
asked Dec 10 '18 at 4:19
Derek Adams
515413
asked Dec 10 '18 at 4:19
Derek Adams
515413
asked Dec 10 '18 at 4:19
Derek Adams
515413
asked Dec 10 '18 at 4:19
Derek Adams
515413
515413
How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information
– user10354138
Dec 10 '18 at 4:27
$G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle.
– Derek Adams
Dec 10 '18 at 5:52
@DerekAdams, did you understand my answer?
– Anubhab Ghosal
Dec 10 '18 at 13:00
add a comment |
How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information
– user10354138
Dec 10 '18 at 4:27
$G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle.
– Derek Adams
Dec 10 '18 at 5:52
@DerekAdams, did you understand my answer?
– Anubhab Ghosal
Dec 10 '18 at 13:00
How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information
– user10354138
Dec 10 '18 at 4:27
How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information
– user10354138
Dec 10 '18 at 4:27
$G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle.
– Derek Adams
Dec 10 '18 at 5:52
$G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle.
– Derek Adams
Dec 10 '18 at 5:52
@DerekAdams, did you understand my answer?
– Anubhab Ghosal
Dec 10 '18 at 13:00
@DerekAdams, did you understand my answer?
– Anubhab Ghosal
Dec 10 '18 at 13:00
add a comment |
1 Answer
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Let $angle CGF=2alpha$, $angle FHE=2beta$ and $angle KCH=gamma$, as shown in the diagram.
$thereforeangle CKH=pi-gamma-beta$ and $angle JGK=frac{pi}{2}-alpha$ .
$angle CDE=pi-gamma-2beta$ and $angle CFG=gamma-2alpha$
Thus, $angle CKH=angle JGK$ if and only if $angle CDE=angle CFG$, whence, exterior angle bisector of $angle CGF$ is parallel to the angle bisector of $angle FHE$, if and only if $CDEF$ is a cyclic quadrilateral.
$blacksquare$
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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active
oldest
votes
Let $angle CGF=2alpha$, $angle FHE=2beta$ and $angle KCH=gamma$, as shown in the diagram.
$thereforeangle CKH=pi-gamma-beta$ and $angle JGK=frac{pi}{2}-alpha$ .
$angle CDE=pi-gamma-2beta$ and $angle CFG=gamma-2alpha$
Thus, $angle CKH=angle JGK$ if and only if $angle CDE=angle CFG$, whence, exterior angle bisector of $angle CGF$ is parallel to the angle bisector of $angle FHE$, if and only if $CDEF$ is a cyclic quadrilateral.
$blacksquare$
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
add a comment |
Let $angle CGF=2alpha$, $angle FHE=2beta$ and $angle KCH=gamma$, as shown in the diagram.
$thereforeangle CKH=pi-gamma-beta$ and $angle JGK=frac{pi}{2}-alpha$ .
$angle CDE=pi-gamma-2beta$ and $angle CFG=gamma-2alpha$
Thus, $angle CKH=angle JGK$ if and only if $angle CDE=angle CFG$, whence, exterior angle bisector of $angle CGF$ is parallel to the angle bisector of $angle FHE$, if and only if $CDEF$ is a cyclic quadrilateral.
$blacksquare$
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
add a comment |
Let $angle CGF=2alpha$, $angle FHE=2beta$ and $angle KCH=gamma$, as shown in the diagram.
$thereforeangle CKH=pi-gamma-beta$ and $angle JGK=frac{pi}{2}-alpha$ .
$angle CDE=pi-gamma-2beta$ and $angle CFG=gamma-2alpha$
Thus, $angle CKH=angle JGK$ if and only if $angle CDE=angle CFG$, whence, exterior angle bisector of $angle CGF$ is parallel to the angle bisector of $angle FHE$, if and only if $CDEF$ is a cyclic quadrilateral.
$blacksquare$
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
Let $angle CGF=2alpha$, $angle FHE=2beta$ and $angle KCH=gamma$, as shown in the diagram.
$thereforeangle CKH=pi-gamma-beta$ and $angle JGK=frac{pi}{2}-alpha$ .
$angle CDE=pi-gamma-2beta$ and $angle CFG=gamma-2alpha$
Thus, $angle CKH=angle JGK$ if and only if $angle CDE=angle CFG$, whence, exterior angle bisector of $angle CGF$ is parallel to the angle bisector of $angle FHE$, if and only if $CDEF$ is a cyclic quadrilateral.
$blacksquare$
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
answered Dec 10 '18 at 6:57
Anubhab Ghosal
81414
81414
add a comment |
add a comment |
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How are the points $D,E$ constructed from $J,G,F,C$? You haven't given us that crucial piece of information
– user10354138
Dec 10 '18 at 4:27
$G$ is the intersection of the extension of sides $DC$ and $FE$, and $H$ is constructed similarly with sides $CF$ and $DE$. $J$ is an arbitrary point on the exterior angle bisector of $angle DGE$. All that is known of $D,E,F,C$ is that they lie on the same circle.
– Derek Adams
Dec 10 '18 at 5:52
@DerekAdams, did you understand my answer?
– Anubhab Ghosal
Dec 10 '18 at 13:00