$AB_1$, $AB_2$, $AB_3$ are the altitude, angle bisector, median from vertex $A$ of $triangle ABC$; arrange...
$begingroup$
Consider an acute angled triangle $triangle ABC$ such that $ABlt AC$.
If from $A$ altitude $AB_1$ is drawn, internal angle bisector $AB_2$ is drawn, and median $AB_3$ is drawn.
Arrange the lengths $BB_1$, $BB_2$ and $BB_3$ in ascending order.
My try: I started with an Isosceles Triangle $triangle ABD$ with $AB=AD$.
Now, for $triangle ABD$, $AB_1$ is altitude, angle bisector, and median.
In figure $2$
Let $angle BAB_1=theta=angle B_1AD$
let $angle DAC=2 beta$
So $angle BAC=2(theta+beta)$
If we construct $AB_2$ asinternal angle bisector of $angle BAC$, Then each half angle is :
$$angle BAB_2=B_2AC=theta+beta gt theta$$
$implies$
$$angle BAB_2 gt angle BAB_1$$
hence the point $B_2$ should be to right side of the point $B_1$
Hence $$BB_1 lt BB_2$$
But can I have a clue to compare $BB_2$ and $BB_3$?
geometry inequality triangles angle
edited Dec 14 '18 at 4:54
Blue
49.5k870158
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
$endgroup$
add a comment |
$begingroup$
Consider an acute angled triangle $triangle ABC$ such that $ABlt AC$.
If from $A$ altitude $AB_1$ is drawn, internal angle bisector $AB_2$ is drawn, and median $AB_3$ is drawn.
Arrange the lengths $BB_1$, $BB_2$ and $BB_3$ in ascending order.
My try: I started with an Isosceles Triangle $triangle ABD$ with $AB=AD$.
Now, for $triangle ABD$, $AB_1$ is altitude, angle bisector, and median.
In figure $2$
Let $angle BAB_1=theta=angle B_1AD$
let $angle DAC=2 beta$
So $angle BAC=2(theta+beta)$
If we construct $AB_2$ asinternal angle bisector of $angle BAC$, Then each half angle is :
$$angle BAB_2=B_2AC=theta+beta gt theta$$
$implies$
$$angle BAB_2 gt angle BAB_1$$
hence the point $B_2$ should be to right side of the point $B_1$
Hence $$BB_1 lt BB_2$$
But can I have a clue to compare $BB_2$ and $BB_3$?
geometry inequality triangles angle
edited Dec 14 '18 at 4:54
Blue
49.5k870158
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
$endgroup$
add a comment |
$begingroup$
Consider an acute angled triangle $triangle ABC$ such that $ABlt AC$.
If from $A$ altitude $AB_1$ is drawn, internal angle bisector $AB_2$ is drawn, and median $AB_3$ is drawn.
Arrange the lengths $BB_1$, $BB_2$ and $BB_3$ in ascending order.
My try: I started with an Isosceles Triangle $triangle ABD$ with $AB=AD$.
Now, for $triangle ABD$, $AB_1$ is altitude, angle bisector, and median.
In figure $2$
Let $angle BAB_1=theta=angle B_1AD$
let $angle DAC=2 beta$
So $angle BAC=2(theta+beta)$
If we construct $AB_2$ asinternal angle bisector of $angle BAC$, Then each half angle is :
$$angle BAB_2=B_2AC=theta+beta gt theta$$
$implies$
$$angle BAB_2 gt angle BAB_1$$
hence the point $B_2$ should be to right side of the point $B_1$
Hence $$BB_1 lt BB_2$$
But can I have a clue to compare $BB_2$ and $BB_3$?
geometry inequality triangles angle
edited Dec 14 '18 at 4:54
Blue
49.5k870158
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
$endgroup$
Consider an acute angled triangle $triangle ABC$ such that $ABlt AC$.
If from $A$ altitude $AB_1$ is drawn, internal angle bisector $AB_2$ is drawn, and median $AB_3$ is drawn.
Arrange the lengths $BB_1$, $BB_2$ and $BB_3$ in ascending order.
My try: I started with an Isosceles Triangle $triangle ABD$ with $AB=AD$.
Now, for $triangle ABD$, $AB_1$ is altitude, angle bisector, and median.
In figure $2$
Let $angle BAB_1=theta=angle B_1AD$
let $angle DAC=2 beta$
So $angle BAC=2(theta+beta)$
If we construct $AB_2$ asinternal angle bisector of $angle BAC$, Then each half angle is :
$$angle BAB_2=B_2AC=theta+beta gt theta$$
$implies$
$$angle BAB_2 gt angle BAB_1$$
hence the point $B_2$ should be to right side of the point $B_1$
Hence $$BB_1 lt BB_2$$
But can I have a clue to compare $BB_2$ and $BB_3$?
geometry inequality triangles angle
geometry inequality triangles angle
edited Dec 14 '18 at 4:54
Blue
49.5k870158
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
edited Dec 14 '18 at 4:54
Blue
49.5k870158
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
edited Dec 14 '18 at 4:54
Blue
49.5k870158
edited Dec 14 '18 at 4:54
Blue
49.5k870158
edited Dec 14 '18 at 4:54
Blue
49.5k870158
49.5k870158
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
asked Dec 14 '18 at 4:13
Umesh shankarUmesh shankar
3,09231220
3,09231220
add a comment |
add a comment |
2 Answers
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$begingroup$
$BB_2:B_2C = AB:AC < 1$ so $BB_2 < BC/2 = BB_3$.
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
We can perceive $ABC$ as a half of a parallelogram $ABDC$ with diagonals $AC, BD.$
Consider a rhombus $ABD'C'$ where $C'in BC$ and $AD',BC'$ are diagonals. Denote $B_1', B_2', B_3'$ the points considered in the question and related to this rhombus.
Diagonals in a rhombus are perpendicular, are angle bisectors of the rhombus, and meet in their common midpoint (as it is for arbitrary parallelogram). Hence the points $B_1', B_2', B_3'$ coincide.
Move $C'$ along $BC$ towards $C$ keeping a parallelogram with the sides $AB;text{and}; AC.$ Clearly, $B_1'$ will not move while $B_2'$ and $B_3'$ do.
Return to the notation $C,B_1,B_2,B_3.$
The angle $angle AB_3C$ becomes obtuse, while $angle BB_3C$ is acute. Consequently, $angle BAB_3 < angle B_3AC.$ Since $AB_2$ is the angle bisector, $B_2$ lies on $BB_3.$
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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active
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active
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votes
$begingroup$
$BB_2:B_2C = AB:AC < 1$ so $BB_2 < BC/2 = BB_3$.
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
$BB_2:B_2C = AB:AC < 1$ so $BB_2 < BC/2 = BB_3$.
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
$endgroup$
add a comment |
$begingroup$
$BB_2:B_2C = AB:AC < 1$ so $BB_2 < BC/2 = BB_3$.
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
$endgroup$
$BB_2:B_2C = AB:AC < 1$ so $BB_2 < BC/2 = BB_3$.
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
answered Dec 14 '18 at 4:44
Quang HoangQuang Hoang
13.2k1233
13.2k1233
add a comment |
add a comment |
$begingroup$
We can perceive $ABC$ as a half of a parallelogram $ABDC$ with diagonals $AC, BD.$
Consider a rhombus $ABD'C'$ where $C'in BC$ and $AD',BC'$ are diagonals. Denote $B_1', B_2', B_3'$ the points considered in the question and related to this rhombus.
Diagonals in a rhombus are perpendicular, are angle bisectors of the rhombus, and meet in their common midpoint (as it is for arbitrary parallelogram). Hence the points $B_1', B_2', B_3'$ coincide.
Move $C'$ along $BC$ towards $C$ keeping a parallelogram with the sides $AB;text{and}; AC.$ Clearly, $B_1'$ will not move while $B_2'$ and $B_3'$ do.
Return to the notation $C,B_1,B_2,B_3.$
The angle $angle AB_3C$ becomes obtuse, while $angle BB_3C$ is acute. Consequently, $angle BAB_3 < angle B_3AC.$ Since $AB_2$ is the angle bisector, $B_2$ lies on $BB_3.$
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
We can perceive $ABC$ as a half of a parallelogram $ABDC$ with diagonals $AC, BD.$
Consider a rhombus $ABD'C'$ where $C'in BC$ and $AD',BC'$ are diagonals. Denote $B_1', B_2', B_3'$ the points considered in the question and related to this rhombus.
Diagonals in a rhombus are perpendicular, are angle bisectors of the rhombus, and meet in their common midpoint (as it is for arbitrary parallelogram). Hence the points $B_1', B_2', B_3'$ coincide.
Move $C'$ along $BC$ towards $C$ keeping a parallelogram with the sides $AB;text{and}; AC.$ Clearly, $B_1'$ will not move while $B_2'$ and $B_3'$ do.
Return to the notation $C,B_1,B_2,B_3.$
The angle $angle AB_3C$ becomes obtuse, while $angle BB_3C$ is acute. Consequently, $angle BAB_3 < angle B_3AC.$ Since $AB_2$ is the angle bisector, $B_2$ lies on $BB_3.$
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
We can perceive $ABC$ as a half of a parallelogram $ABDC$ with diagonals $AC, BD.$
Consider a rhombus $ABD'C'$ where $C'in BC$ and $AD',BC'$ are diagonals. Denote $B_1', B_2', B_3'$ the points considered in the question and related to this rhombus.
Diagonals in a rhombus are perpendicular, are angle bisectors of the rhombus, and meet in their common midpoint (as it is for arbitrary parallelogram). Hence the points $B_1', B_2', B_3'$ coincide.
Move $C'$ along $BC$ towards $C$ keeping a parallelogram with the sides $AB;text{and}; AC.$ Clearly, $B_1'$ will not move while $B_2'$ and $B_3'$ do.
Return to the notation $C,B_1,B_2,B_3.$
The angle $angle AB_3C$ becomes obtuse, while $angle BB_3C$ is acute. Consequently, $angle BAB_3 < angle B_3AC.$ Since $AB_2$ is the angle bisector, $B_2$ lies on $BB_3.$
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
$endgroup$
We can perceive $ABC$ as a half of a parallelogram $ABDC$ with diagonals $AC, BD.$
Consider a rhombus $ABD'C'$ where $C'in BC$ and $AD',BC'$ are diagonals. Denote $B_1', B_2', B_3'$ the points considered in the question and related to this rhombus.
Diagonals in a rhombus are perpendicular, are angle bisectors of the rhombus, and meet in their common midpoint (as it is for arbitrary parallelogram). Hence the points $B_1', B_2', B_3'$ coincide.
Move $C'$ along $BC$ towards $C$ keeping a parallelogram with the sides $AB;text{and}; AC.$ Clearly, $B_1'$ will not move while $B_2'$ and $B_3'$ do.
Return to the notation $C,B_1,B_2,B_3.$
The angle $angle AB_3C$ becomes obtuse, while $angle BB_3C$ is acute. Consequently, $angle BAB_3 < angle B_3AC.$ Since $AB_2$ is the angle bisector, $B_2$ lies on $BB_3.$
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
edited Dec 31 '18 at 11:53
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
answered Dec 14 '18 at 17:47
user376343user376343
3,9584829
3,9584829
add a comment |
add a comment |
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