If AH and BG are angle bisectors, how would I find IJ?

Question

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 $$GB=\dfrac{4\sqrt{10}}{3}$$ I really don't know what to do from here, could someone help me?

Answer 1

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:

$$\frac{3\cdot0+4\cdot3 + 5\cdot0}{12}$$

Answer 2

Knowing $AG$, you can figure out $BI : IG$ by the angle bisector theorem. Then use $\triangle BIJ \sim \triangle BGC$.

Answer 3

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 $$r=1.$$

Answer 4

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)