Circle inscribed in an angle, simpler proof needed 
Circle $C$ is inscribed in angle $\angle PAQ$. Line $QS$ and leg $AP$ are parallel. Line $AS$ intersects circle at point $R$. Line $QR$ intersects leg $AP$ at point $T$. Prove that $AT=TP$.
Effort so far: Actually, I was able to solve this problem but the solution seems too complicated to me. Line $PQ$ crosses line $AS$ at point $U$. You can show that point $U$ is a harmonic conjugate of $A$ with respect to $R$, $S$. By using that fact and some more projective geometry I was able to prove that $AT=TP$.
But I still believe that an "elementary" proof (one that is not based on projective geometry) is still available. Are you able to find such one?
 A: 
$\angle AQT = \angle QSA = \angle RAT$
and $\angle QTA= \angle RTA$
$$\implies \Delta QTA \sim \Delta ATR$$
$$\therefore \frac {TQ}{AT}= \frac {AT}{TR}$$
$$TR \cdot TQ = AT^2    \tag{1}$$
On the other hand, $TP$ is tangent to the circle implies that
$$TR \cdot TQ = TP^2    \tag{2}$$
$(1)$ and $(2) \implies$
$$AT=TP$$
A: First of all, it is very easy to see that we only need to prove $\frac{RS}{QR}=2$ because $\triangle ATQ$ and $\triangle QRS$ are similar.
To prove that, let's assume: $\angle QAS=\theta_1, \angle PAS=\theta_2, \angle QAP=A^{\circ}$, and the radius of the circle is $R.$
Then it is easy to verify that $\angle QCS= 2A^{\circ}$, therefore we have:
$$QS=2R\sin A^{\circ} \\ AQ=R \cot \frac{A^{\circ}}{2}. $$
Hence:
$$\frac{QS}{AQ}=\frac{\sin \theta_1}{\sin \theta_2}=4\sin^2 \frac{A^{\circ}}{2} \implies \frac{\sin (A^{\circ}-\theta_2)}{\sin \theta_2} =4\sin^2 \frac{A^{\circ}}{2}.$$
On the other hand: $$\frac{RS}{QR}=\frac{\sin (2\theta_2+\theta_1)}{\sin \theta_2}=\frac{\sin (A^{\circ}+\theta_2)}{\sin \theta_2}.$$
Now, observe that:
$$\frac{RS}{QR}-\frac{QS}{AQ}=\frac{\sin (A^{\circ}+\theta_2)}{\sin \theta_2}-\frac{\sin (A^{\circ}-\theta_2)}{\sin \theta_2}=2\cos A^{\circ}\\ \implies \frac{RS}{QR}=2\cos A^{\circ}+4\sin^2 \frac{A^{\circ}}{2}=2.$$
So, we are done.
A: Hint:
Draw diameter PB passing center C. Draw a line from P parallel with TB. it intersects tangent on circle at B at point F.Connect A to B. It is clear that $BF=TP$. Connect T to F , it intersects circle at points D(close to T) and G(close to B). Now if we show $AB||TF$, then we we conclude that $AT=BF=TP$. For this Mark intersection of AB and circle as E. Connect C to E. You have to show that $CE\bot TE$ and get result that $\triangle CBF=\triangle CET$ and:
$\angle ECT=\angle BCF$
$\Rightarrow \overset{\large\frown}{ED}=\overset{\large\frown}{BG}$
which means $BE||TE$.
Update: To show $CE\bot TE$ , extend BF, it intersect extension of AC at H. Draw a line from H parallel with AQ, it touches circle at tangent point Q', so QQ' is a diameter of the circle. The construction is symmetric about center C. Draw a line from H to P, it intersects the circle at E'. E' is mirror of E about C, so EE' is another diameter of the circle, that means $CE\bot TE$.
