Mixing
A conjecture involving angle bisectors
Clash Royale CLAN TAG#URR8PPP
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Consider an equilateral triangle $ADB$, and draw any triangle $ACB$, where $C$ is contained in the equilateral triangle $ADB$. 
We draw two circles: one with center in $A$ and passing by $C$ (obtaining the point $K$ on the side $AB$ and the point $G$ on the side $AD$); the other one with center in $B$ and passing by $C$ (obtaining the point $J$ on the side $AB$ and the point $H$ on the side $DB$).
The side $AB$ results subdivided in three segments $AJ$ (blue), $JK$ (red), and $KB$ (green).
Now, we draw two other circles: one with center in $A$ and passing by $J$ (obtaining the point $F$ on the side $AD$ and the point $M$ on the side $AC$), and the other one with center in $B$ and passing by $K$ (obtaining the point $I$ on the side $DB$ and the point $N$ on the side $CB$).
This way, also the other sides $AD$ and $DB$ of the equilateral triangles result subdivided in three segments.
Let us now draw the angle bisectors of $F,B,G$ (green) and of $H,A,I$ (blue), intersecting in the point $L$.
My conjecture is that, no matter the exact position of $C$, the angle bisectors of $F,B,G$ and of $H,A,I$ form always and angle equal to $frac2pi3$.
Moreover, the point $L$ always lies on an arc of circle (whose center is located on the specular point of $D$ with respect to the side $AB$).
It is probably a very obvious result, and I apologize for the naivety in this case. However,
Is there an elementary proof for such conjecture?
Thanks for your help and for your suggestions!
geometry euclidean-geometry triangle
asked Jul 18 at 22:49
Andrea Prunotto
578215
add a comment |Â
up vote
2
down vote
favorite
Consider an equilateral triangle $ADB$, and draw any triangle $ACB$, where $C$ is contained in the equilateral triangle $ADB$.
We draw two circles: one with center in $A$ and passing by $C$ (obtaining the point $K$ on the side $AB$ and the point $G$ on the side $AD$); the other one with center in $B$ and passing by $C$ (obtaining the point $J$ on the side $AB$ and the point $H$ on the side $DB$).
The side $AB$ results subdivided in three segments $AJ$ (blue), $JK$ (red), and $KB$ (green).
Now, we draw two other circles: one with center in $A$ and passing by $J$ (obtaining the point $F$ on the side $AD$ and the point $M$ on the side $AC$), and the other one with center in $B$ and passing by $K$ (obtaining the point $I$ on the side $DB$ and the point $N$ on the side $CB$).
This way, also the other sides $AD$ and $DB$ of the equilateral triangles result subdivided in three segments.
Let us now draw the angle bisectors of $F,B,G$ (green) and of $H,A,I$ (blue), intersecting in the point $L$.
My conjecture is that, no matter the exact position of $C$, the angle bisectors of $F,B,G$ and of $H,A,I$ form always and angle equal to $frac2pi3$.
Moreover, the point $L$ always lies on an arc of circle (whose center is located on the specular point of $D$ with respect to the side $AB$).
It is probably a very obvious result, and I apologize for the naivety in this case. However,
Is there an elementary proof for such conjecture?
Thanks for your help and for your suggestions!
geometry euclidean-geometry triangle
asked Jul 18 at 22:49
Andrea Prunotto
578215
1
Rotating $triangle ABD$ by $120^circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$.
– Blue
Jul 18 at 23:44
1
BTW, here's an equivalent construction that avoids the point $C$: Let $overlineJK$ be any subsegment of $overlineAB$ in equilateral $triangle ABD$. Transfer the endpoints of that segment onto $overlineAD$ and $overlineBD$ via circles about $A$ and $B$; this gives your segments $overlineFG$ and $overlineHI$. Then, the bisectors of $angle FBG$ and $angle HAI$ make an angle of $120^circ$.
– Blue
Jul 18 at 23:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider an equilateral triangle $ADB$, and draw any triangle $ACB$, where $C$ is contained in the equilateral triangle $ADB$.
We draw two circles: one with center in $A$ and passing by $C$ (obtaining the point $K$ on the side $AB$ and the point $G$ on the side $AD$); the other one with center in $B$ and passing by $C$ (obtaining the point $J$ on the side $AB$ and the point $H$ on the side $DB$).
The side $AB$ results subdivided in three segments $AJ$ (blue), $JK$ (red), and $KB$ (green).
Now, we draw two other circles: one with center in $A$ and passing by $J$ (obtaining the point $F$ on the side $AD$ and the point $M$ on the side $AC$), and the other one with center in $B$ and passing by $K$ (obtaining the point $I$ on the side $DB$ and the point $N$ on the side $CB$).
This way, also the other sides $AD$ and $DB$ of the equilateral triangles result subdivided in three segments.
Let us now draw the angle bisectors of $F,B,G$ (green) and of $H,A,I$ (blue), intersecting in the point $L$.
My conjecture is that, no matter the exact position of $C$, the angle bisectors of $F,B,G$ and of $H,A,I$ form always and angle equal to $frac2pi3$.
Moreover, the point $L$ always lies on an arc of circle (whose center is located on the specular point of $D$ with respect to the side $AB$).
It is probably a very obvious result, and I apologize for the naivety in this case. However,
Is there an elementary proof for such conjecture?
Thanks for your help and for your suggestions!
geometry euclidean-geometry triangle
asked Jul 18 at 22:49
Andrea Prunotto
578215
Consider an equilateral triangle $ADB$, and draw any triangle $ACB$, where $C$ is contained in the equilateral triangle $ADB$.
We draw two circles: one with center in $A$ and passing by $C$ (obtaining the point $K$ on the side $AB$ and the point $G$ on the side $AD$); the other one with center in $B$ and passing by $C$ (obtaining the point $J$ on the side $AB$ and the point $H$ on the side $DB$).
The side $AB$ results subdivided in three segments $AJ$ (blue), $JK$ (red), and $KB$ (green).
Now, we draw two other circles: one with center in $A$ and passing by $J$ (obtaining the point $F$ on the side $AD$ and the point $M$ on the side $AC$), and the other one with center in $B$ and passing by $K$ (obtaining the point $I$ on the side $DB$ and the point $N$ on the side $CB$).
This way, also the other sides $AD$ and $DB$ of the equilateral triangles result subdivided in three segments.
Let us now draw the angle bisectors of $F,B,G$ (green) and of $H,A,I$ (blue), intersecting in the point $L$.
My conjecture is that, no matter the exact position of $C$, the angle bisectors of $F,B,G$ and of $H,A,I$ form always and angle equal to $frac2pi3$.
Moreover, the point $L$ always lies on an arc of circle (whose center is located on the specular point of $D$ with respect to the side $AB$).
It is probably a very obvious result, and I apologize for the naivety in this case. However,
Is there an elementary proof for such conjecture?
Thanks for your help and for your suggestions!
geometry euclidean-geometry triangle
asked Jul 18 at 22:49
Andrea Prunotto
578215
edited Jul 18 at 23:10
asked Jul 18 at 22:49
Andrea Prunotto
578215
asked Jul 18 at 22:49
Andrea Prunotto
578215
asked Jul 18 at 22:49
Andrea Prunotto
578215
578215
1
Rotating $triangle ABD$ by $120^circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$.
– Blue
Jul 18 at 23:44
1
BTW, here's an equivalent construction that avoids the point $C$: Let $overlineJK$ be any subsegment of $overlineAB$ in equilateral $triangle ABD$. Transfer the endpoints of that segment onto $overlineAD$ and $overlineBD$ via circles about $A$ and $B$; this gives your segments $overlineFG$ and $overlineHI$. Then, the bisectors of $angle FBG$ and $angle HAI$ make an angle of $120^circ$.
– Blue
Jul 18 at 23:55
add a comment |Â
1
Rotating $triangle ABD$ by $120^circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$.
– Blue
Jul 18 at 23:44
1
BTW, here's an equivalent construction that avoids the point $C$: Let $overlineJK$ be any subsegment of $overlineAB$ in equilateral $triangle ABD$. Transfer the endpoints of that segment onto $overlineAD$ and $overlineBD$ via circles about $A$ and $B$; this gives your segments $overlineFG$ and $overlineHI$. Then, the bisectors of $angle FBG$ and $angle HAI$ make an angle of $120^circ$.
– Blue
Jul 18 at 23:55
1
1
Rotating $triangle ABD$ by $120^circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$.
– Blue
Jul 18 at 23:44
Rotating $triangle ABD$ by $120^circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$.
– Blue
Jul 18 at 23:44
1
1
BTW, here's an equivalent construction that avoids the point $C$: Let $overlineJK$ be any subsegment of $overlineAB$ in equilateral $triangle ABD$. Transfer the endpoints of that segment onto $overlineAD$ and $overlineBD$ via circles about $A$ and $B$; this gives your segments $overlineFG$ and $overlineHI$. Then, the bisectors of $angle FBG$ and $angle HAI$ make an angle of $120^circ$.
– Blue
Jul 18 at 23:55
BTW, here's an equivalent construction that avoids the point $C$: Let $overlineJK$ be any subsegment of $overlineAB$ in equilateral $triangle ABD$. Transfer the endpoints of that segment onto $overlineAD$ and $overlineBD$ via circles about $A$ and $B$; this gives your segments $overlineFG$ and $overlineHI$. Then, the bisectors of $angle FBG$ and $angle HAI$ make an angle of $120^circ$.
– Blue
Jul 18 at 23:55
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
As it seems you have noted by the coloring of your segments,
$$AF=DH,FG=HI,GD=IB.$$
We then see that, as (degenerate) pentagons,
$$BAFGDsim ADHIB.$$
Let $L$ be the intersection of the bisectors angles $HAI$ and $FBG$. We then have
$$angle LAB=angle IAB + fracangle HAI2=angle GBD+fracangle FBG2=angle LBD$$
(where the second equality is by our similarity), so
$$angle LAB=60^circ-angle LBA implies angle BLA=120^circ.$$
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As it seems you have noted by the coloring of your segments,
$$AF=DH,FG=HI,GD=IB.$$
We then see that, as (degenerate) pentagons,
$$BAFGDsim ADHIB.$$
Let $L$ be the intersection of the bisectors angles $HAI$ and $FBG$. We then have
$$angle LAB=angle IAB + fracangle HAI2=angle GBD+fracangle FBG2=angle LBD$$
(where the second equality is by our similarity), so
$$angle LAB=60^circ-angle LBA implies angle BLA=120^circ.$$
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
add a comment |Â
up vote
1
down vote
accepted
As it seems you have noted by the coloring of your segments,
$$AF=DH,FG=HI,GD=IB.$$
We then see that, as (degenerate) pentagons,
$$BAFGDsim ADHIB.$$
Let $L$ be the intersection of the bisectors angles $HAI$ and $FBG$. We then have
$$angle LAB=angle IAB + fracangle HAI2=angle GBD+fracangle FBG2=angle LBD$$
(where the second equality is by our similarity), so
$$angle LAB=60^circ-angle LBA implies angle BLA=120^circ.$$
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As it seems you have noted by the coloring of your segments,
$$AF=DH,FG=HI,GD=IB.$$
We then see that, as (degenerate) pentagons,
$$BAFGDsim ADHIB.$$
Let $L$ be the intersection of the bisectors angles $HAI$ and $FBG$. We then have
$$angle LAB=angle IAB + fracangle HAI2=angle GBD+fracangle FBG2=angle LBD$$
(where the second equality is by our similarity), so
$$angle LAB=60^circ-angle LBA implies angle BLA=120^circ.$$
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
As it seems you have noted by the coloring of your segments,
$$AF=DH,FG=HI,GD=IB.$$
We then see that, as (degenerate) pentagons,
$$BAFGDsim ADHIB.$$
Let $L$ be the intersection of the bisectors angles $HAI$ and $FBG$. We then have
$$angle LAB=angle IAB + fracangle HAI2=angle GBD+fracangle FBG2=angle LBD$$
(where the second equality is by our similarity), so
$$angle LAB=60^circ-angle LBA implies angle BLA=120^circ.$$
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
answered Jul 18 at 23:26
Carl Schildkraut
8,26211238
8,26211238
add a comment |Â
add a comment |Â
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1
Rotating $triangle ABD$ by $120^circ$ (clockwise) about its center carries the bisector through $B$ onto the (original) bisector through $A$.
– Blue
Jul 18 at 23:44
1
BTW, here's an equivalent construction that avoids the point $C$: Let $overlineJK$ be any subsegment of $overlineAB$ in equilateral $triangle ABD$. Transfer the endpoints of that segment onto $overlineAD$ and $overlineBD$ via circles about $A$ and $B$; this gives your segments $overlineFG$ and $overlineHI$. Then, the bisectors of $angle FBG$ and $angle HAI$ make an angle of $120^circ$.
– Blue
Jul 18 at 23:55