# $ABC$ exists so $\overline{AX}$, $\overline{BC}$, and $\overline{CZ}$ are concurrent. $\overline{AB} = 6$, $\overline{AC} = 3$, $\overline {BX}$ is?

A certain triangle $$ABC$$ exists such that it's angle bisector $$\overline{AX}$$, median $$\overline{BC}$$, and altitude $$\overline{CZ}$$ are concurrent. If the lengths $$\overline{AB} = 6$$ and $$\overline{AC} = 3$$, then what is the length of line $$\overline {BX}$$?

Since the angle bisector, median, and altitude are all cevians of our triangle $$ABC$$, I figured that Ceva's theorem might help here. But after this, I'm not sure how to proceed. I've never seen a problem combine all three of these unique cevains and try to get the length of one of them. Does anyone have any ideas?

Ceva gives $$BX\cdot CY\cdot AZ=BZ\cdot AY\cdot CX$$ and since $$AY=CY$$ and $$BX=2CX$$, we obtain: $$2AZ=BZ$$ or in the standard notation $$2\cdot b\cdot\frac{b^2+c^2-a^2}{2bc}=a\cdot\frac{a^2+c^2-b^2}{2ac}.$$ From here we can find a value of $$a$$ and after this $$BX$$.