In a triangle $ABC$, the median $AM$, the interior bisector CN and the cevian $BD$ are drawn concurrently. $NT \perp AC$ is plotted. Find $TS$ if $CD = 6$ and $S = AM \cap ND$ (answer:3)
My progress: Here's the drawing (in scale) and the relations I found... I couldn't use the $NT$ perpendicular.
$BM=CM$
By Ceva's theorem.: $6\cdot AN \cdot x =AD \cdot BN \cdot x\\ \therefore \boxed{6\cdot AN = AD\cdot BN} \\$
From Angle Bissector Theorem: $\boxed{\frac{BN}{2x} = \frac{AN}{6+AD}}\\$
Also from Menelaus's Theorem to $\triangle BDC$ with transversal $AM\\$: $x\cdot BF \cdot AD = x \cdot DF \cdot (6+AD)\\ \therefore \boxed{BF\cdot AD =DF(6+AD)}\\ $
From Menelaus's Theorem to $\triangle AMC-BD:\\$ $6 \cdot AF\cdot x = AD \cdot FM \cdot 2x \\ \therefore \boxed{3AF = AD\cdot FM}\\$
Menelaus Theorem to $\triangle ANC - BD:\\$ $\boxed{AD \cdot CF \cdot BN = 6 \cdot FN \cdot AB}$
By Ceva's Theorem,
$CD\cdot AN \cdot x = DA \cdot NB \cdot x \implies \frac{AD}{DC} = \frac{AN}{NB}$
So by Triangle Proportionality Theorem Converse, $DN \parallel CB$ and therefore $\angle DNC = \theta$.
That leads to $DN = DC = 6$
As $AM$ bisects $CB$, it also bisects $DN$. As $S$ is the midpoint of hypotenuse $DN$ in right triangle $\triangle DTN$, $S$ must be the circumcenter. So it follows that $ST = DS = SN = 3$.