A Question regarding radius of circumcircle and sides of a triangle NOTE: I am looking for a hint,not the whole solution.
A question from BdMO Nationals 2012

Given triangle $ABC$, the square $PQRS$ is drawn such that $P$,$Q$ are on BC, $R$ is on $CA$ and $S$ on $AB$. Radius of the circle that passes through $A, B, C$ is $T$. If $AB = c, BC = a, CA = b$,show that $\dfrac{AS}{BS}=\dfrac{bc}{2T\cdot BC}$

A picture would help here.It is easy to see that $SR$ is parallel to BC.Therefore,
$\dfrac{AS}{BS}=\dfrac{AR}{RC}$
We note that the left side of the equality is also the left side of the equality to be proved.Therefore,we actually need to prove that 
$\dfrac{AR}{RC}=\dfrac{bc}{2T\cdot BC}$
I am lost at this point.However,I can see that we need to use T somewhere.However,I can't relate T to any side of the square or the triangle.So I need some help doing that.A small(teeny-tiny if possible) hint would be appreciated.
NOTE: No trigonometry is allowed.
 A: Hint:
We need to use $T$, so why not draw a circle! 
$O$ is the center of the circle. $AH\perp BC$.
$\frac{AS}{BS}=\frac{AH'}{SP}=\frac{AH'}{RS}=\frac{AH}{BC}$
$\triangle ABF \sim \triangle AHC$

A: Consider the height of the triangle above the square, $h$, and the side of the square, $s$:
$\hspace{3.8cm}$
Note that by similar triangles,
$$
\frac{AS}{BS}=\frac hs=\frac{h+s}{a}\tag{1}
$$
We also have a couple of formulas involving the radius of the circumcircle, $T$, and the area of the triangle, $E$:
$$
abc=4TE\tag{2}
$$
and
$$
2E=(h+s)a\tag{3}
$$
A bit of algebra should get what you want.
A: first consider a $\bigtriangleup$ABC with circumcircle as shown in fig:1


*

*AO is diametre and AD is altitude.we name AD=h ;AO=2r;AB=c;AC=b;BC=a

*so $\bigtriangleup$ADB~$\bigtriangleup$AOC. so $\frac{2r}{c}$=$\frac{b}{h}$

*so bc=2rh so abc=2rha or we have r=$\frac{abc}{4\Delta}$[here r=your T][also $\frac{1}{2}$ah=area]

*so putting the value in the above equation we have this to prove-$\frac{AS}{BS}$=$\frac{h}{a}$

*see $\Delta$ABD~$\Delta$BSP so $\frac{AB}{BS}$=$\frac{h}{SP}$

*and $\frac{AS}{AB}$=$\frac{SR}{a}$[$\Delta$ASR~$\Delta$ABC] multiplying the two equations we get -$\frac{AS}{BS}$=$\frac{h}{a}$[proved]
A: use the area formula :  $\Delta=abc/4R$
$\dfrac{AR}{RC}=\dfrac{bc}{2T\cdot BC}$$\Longrightarrow$$\dfrac{AS}{BS}=\dfrac{h}{a}$$\Longrightarrow$$\dfrac{AS}{AB}=\dfrac{AR}{AC}=\dfrac{h}{a+h}$$\Longrightarrow$
$(\dfrac{h}{a+h})^2=\dfrac{h-t}{h}\cdot\dfrac{t}{a}$$\Longrightarrow$$(\dfrac{h}{a+h})^2=h^2\cdot{\dfrac{h}{a+h}}\cdot{\dfrac{a}{a+h}}\cdot{\dfrac{1}{ah}}=(\dfrac{h}{a+h})^2$
so, your solution holds !
