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

Question

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?

Answer 1

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 $BX$.

Can you end it now?