If AH and BG are angle bisectors, how would I find IJ?
$begingroup$
Diagram
I've tried finding it, but it just doesn't seem to come out.
I found $$GC=dfrac{4}{3}$$ $$AG=dfrac{5}{3}$$ and $$GB=dfrac{4sqrt{10}}{3}$$ I really don't know what to do from here, could someone help me?
geometry euclidean-geometry triangles
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
asked Jan 15 at 0:00
weareallinweareallin
8418
$endgroup$
add a comment |
$begingroup$
Diagram
I've tried finding it, but it just doesn't seem to come out.
I found $$GC=dfrac{4}{3}$$ $$AG=dfrac{5}{3}$$ and $$GB=dfrac{4sqrt{10}}{3}$$ I really don't know what to do from here, could someone help me?
geometry euclidean-geometry triangles
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
asked Jan 15 at 0:00
weareallinweareallin
8418
$endgroup$
$begingroup$
Do you know trigonometry?
$endgroup$
– Clayton
Jan 15 at 0:15
$begingroup$
No, I haven't gotten to it yet. I'm wondering if this is solvable without trigonometry.
$endgroup$
– weareallin
Jan 15 at 0:18
add a comment |
$begingroup$
Diagram
I've tried finding it, but it just doesn't seem to come out.
I found $$GC=dfrac{4}{3}$$ $$AG=dfrac{5}{3}$$ and $$GB=dfrac{4sqrt{10}}{3}$$ I really don't know what to do from here, could someone help me?
geometry euclidean-geometry triangles
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
asked Jan 15 at 0:00
weareallinweareallin
8418
$endgroup$
Diagram
I've tried finding it, but it just doesn't seem to come out.
I found $$GC=dfrac{4}{3}$$ $$AG=dfrac{5}{3}$$ and $$GB=dfrac{4sqrt{10}}{3}$$ I really don't know what to do from here, could someone help me?
geometry euclidean-geometry triangles
geometry euclidean-geometry triangles
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
asked Jan 15 at 0:00
weareallinweareallin
8418
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
asked Jan 15 at 0:00
weareallinweareallin
8418
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
edited Jan 15 at 6:28
Michael Rozenberg
111k1897201
111k1897201
asked Jan 15 at 0:00
weareallinweareallin
8418
asked Jan 15 at 0:00
weareallinweareallin
8418
asked Jan 15 at 0:00
weareallinweareallin
8418
8418
$begingroup$
Do you know trigonometry?
$endgroup$
– Clayton
Jan 15 at 0:15
$begingroup$
No, I haven't gotten to it yet. I'm wondering if this is solvable without trigonometry.
$endgroup$
– weareallin
Jan 15 at 0:18
add a comment |
$begingroup$
Do you know trigonometry?
$endgroup$
– Clayton
Jan 15 at 0:15
$begingroup$
No, I haven't gotten to it yet. I'm wondering if this is solvable without trigonometry.
$endgroup$
– weareallin
Jan 15 at 0:18
$begingroup$
Do you know trigonometry?
$endgroup$
– Clayton
Jan 15 at 0:15
$begingroup$
Do you know trigonometry?
$endgroup$
– Clayton
Jan 15 at 0:15
$begingroup$
No, I haven't gotten to it yet. I'm wondering if this is solvable without trigonometry.
$endgroup$
– weareallin
Jan 15 at 0:18
$begingroup$
No, I haven't gotten to it yet. I'm wondering if this is solvable without trigonometry.
$endgroup$
– weareallin
Jan 15 at 0:18
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
Point $I$ is the incenter of $Delta ABC$. Let $B = (0,0)$, $C = (4,0)$ and $A = (4,3)$. Then, the y-coordinate of $I$ is:
$$frac{3cdot0+4cdot3 + 5cdot0}{12}$$
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
$endgroup$
add a comment |
$begingroup$
Knowing $AG$, you can figure out $BI : IG$ by the angle bisector theorem. Then use $triangle BIJ sim triangle BGC$.
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
$endgroup$
add a comment |
$begingroup$
Let $IJ=r$, $IK$ be a perpendicular from $I$ to $AC$ and $IL$ be a perpendicular from $I$ to $AB.$
Thus, since $IJCK$ is square, we obtain
$$AL+BL=AB$$ or
$$3-r+4-r=5,$$ which gives $$r=1.$$
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
If you know area of a triangle is also:
$Delta=rs$
Where $r$ is inradius and $s$ is semiperimeter.
$Delta$ = 1/2 × base × height
Form the equation and you will get $IJ$.
$IJ$ = 1
(If you need to cross check)
answered Jan 17 at 11:31
RB MCPERB MCPE
262
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Point $I$ is the incenter of $Delta ABC$. Let $B = (0,0)$, $C = (4,0)$ and $A = (4,3)$. Then, the y-coordinate of $I$ is:
$$frac{3cdot0+4cdot3 + 5cdot0}{12}$$
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
$endgroup$
add a comment |
$begingroup$
Hint:
Point $I$ is the incenter of $Delta ABC$. Let $B = (0,0)$, $C = (4,0)$ and $A = (4,3)$. Then, the y-coordinate of $I$ is:
$$frac{3cdot0+4cdot3 + 5cdot0}{12}$$
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
$endgroup$
add a comment |
$begingroup$
Hint:
Point $I$ is the incenter of $Delta ABC$. Let $B = (0,0)$, $C = (4,0)$ and $A = (4,3)$. Then, the y-coordinate of $I$ is:
$$frac{3cdot0+4cdot3 + 5cdot0}{12}$$
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
$endgroup$
Hint:
Point $I$ is the incenter of $Delta ABC$. Let $B = (0,0)$, $C = (4,0)$ and $A = (4,3)$. Then, the y-coordinate of $I$ is:
$$frac{3cdot0+4cdot3 + 5cdot0}{12}$$
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
answered Jan 15 at 0:33
Toby MakToby Mak
3,69011128
3,69011128
add a comment |
add a comment |
$begingroup$
Knowing $AG$, you can figure out $BI : IG$ by the angle bisector theorem. Then use $triangle BIJ sim triangle BGC$.
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
$endgroup$
add a comment |
$begingroup$
Knowing $AG$, you can figure out $BI : IG$ by the angle bisector theorem. Then use $triangle BIJ sim triangle BGC$.
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
$endgroup$
add a comment |
$begingroup$
Knowing $AG$, you can figure out $BI : IG$ by the angle bisector theorem. Then use $triangle BIJ sim triangle BGC$.
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
$endgroup$
Knowing $AG$, you can figure out $BI : IG$ by the angle bisector theorem. Then use $triangle BIJ sim triangle BGC$.
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
answered Jan 15 at 0:43
Misha LavrovMisha Lavrov
49.9k759110
49.9k759110
add a comment |
add a comment |
$begingroup$
Let $IJ=r$, $IK$ be a perpendicular from $I$ to $AC$ and $IL$ be a perpendicular from $I$ to $AB.$
Thus, since $IJCK$ is square, we obtain
$$AL+BL=AB$$ or
$$3-r+4-r=5,$$ which gives $$r=1.$$
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
Let $IJ=r$, $IK$ be a perpendicular from $I$ to $AC$ and $IL$ be a perpendicular from $I$ to $AB.$
Thus, since $IJCK$ is square, we obtain
$$AL+BL=AB$$ or
$$3-r+4-r=5,$$ which gives $$r=1.$$
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
add a comment |
$begingroup$
Let $IJ=r$, $IK$ be a perpendicular from $I$ to $AC$ and $IL$ be a perpendicular from $I$ to $AB.$
Thus, since $IJCK$ is square, we obtain
$$AL+BL=AB$$ or
$$3-r+4-r=5,$$ which gives $$r=1.$$
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
$endgroup$
Let $IJ=r$, $IK$ be a perpendicular from $I$ to $AC$ and $IL$ be a perpendicular from $I$ to $AB.$
Thus, since $IJCK$ is square, we obtain
$$AL+BL=AB$$ or
$$3-r+4-r=5,$$ which gives $$r=1.$$
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
answered Jan 15 at 6:25
Michael RozenbergMichael Rozenberg
111k1897201
111k1897201
add a comment |
add a comment |
$begingroup$
If you know area of a triangle is also:
$Delta=rs$
Where $r$ is inradius and $s$ is semiperimeter.
$Delta$ = 1/2 × base × height
Form the equation and you will get $IJ$.
$IJ$ = 1
(If you need to cross check)
answered Jan 17 at 11:31
RB MCPERB MCPE
262
$endgroup$
add a comment |
$begingroup$
If you know area of a triangle is also:
$Delta=rs$
Where $r$ is inradius and $s$ is semiperimeter.
$Delta$ = 1/2 × base × height
Form the equation and you will get $IJ$.
$IJ$ = 1
(If you need to cross check)
answered Jan 17 at 11:31
RB MCPERB MCPE
262
$endgroup$
add a comment |
$begingroup$
If you know area of a triangle is also:
$Delta=rs$
Where $r$ is inradius and $s$ is semiperimeter.
$Delta$ = 1/2 × base × height
Form the equation and you will get $IJ$.
$IJ$ = 1
(If you need to cross check)
answered Jan 17 at 11:31
RB MCPERB MCPE
262
$endgroup$
If you know area of a triangle is also:
$Delta=rs$
Where $r$ is inradius and $s$ is semiperimeter.
$Delta$ = 1/2 × base × height
Form the equation and you will get $IJ$.
$IJ$ = 1
(If you need to cross check)
answered Jan 17 at 11:31
RB MCPERB MCPE
262
edited Jan 17 at 13:03
answered Jan 17 at 11:31
RB MCPERB MCPE
262
answered Jan 17 at 11:31
RB MCPERB MCPE
262
answered Jan 17 at 11:31
RB MCPERB MCPE
262
262
add a comment |
add a comment |
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$begingroup$
Do you know trigonometry?
$endgroup$
– Clayton
Jan 15 at 0:15
$begingroup$
No, I haven't gotten to it yet. I'm wondering if this is solvable without trigonometry.
$endgroup$
– weareallin
Jan 15 at 0:18