A geometry problem by A Former Brilliant Member

Apply pythagorean theorem in $\triangle ABC$ ,

$BC^2=25^2-20^2$ $\implies$ $BC^2=225$ $\implies$ $BC=\sqrt{225}$ $\implies$ $BC=15$

It follows that, $BP=15-5=10$ .

In $\triangle ABC$ , since $AP,BQ$ and $CS$ are concurrent, we have

$\dfrac{AQ}{QC} \times \dfrac{CP}{PB} \times \dfrac{BS}{SA}=1$ ( Ceva's Theorem )

Substituting, we have

$\dfrac{15}{5} \times \dfrac{5}{10} \times \dfrac{BS}{25-BS}=1$ $\implies$ $BS=10$

Now, consider $\triangle ABC$ with transversal $QPT$ .

$\dfrac{AQ}{QC} \times \dfrac{CP}{PB} \times \dfrac{BT}{TA}=-1$ ( Menelaus' Theorem )

Since we are not dealing with directed line segments, this may be restated as

$(AQ)(CP)(BT)=(QC)(PB)(AT)$

Substituting, we have

$(15)(5)(BT)=(5)(10)(B+25)$ $\implies$ $BT=50$

Finally, $TS=BT+BS=50+10=$ $\color{#3D99F6}\large\boxed{60}$ $\boxed{\text{answer}}$

I used Menelaus' Theorem on four triangles after naming intersection of CS and PQ as V.
$\Delta s~and ~transversals:- BSC~with~TPV, ~~~and ~SAC~with~TVQ~ ~~and~got ~AT/TB=15/10\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ~~~~~~~~~~ BSC~with~PRA, ~~~and ~SAC~with~BRQ~~~ and~got ~BS/AS=10/15\\ And~used ~componendo-dividendo~ to~ obtain~~TB=50~and~ BS=10.~~TS=60.$
Brilliant Member's solution is a better one.