I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
$\underline{\bf{\color{red}{Note}}}$:$\space$
We provide below an answer/proof. But we have a request. Before going through this answer, you need to finish the conversation you have started with @Lynn to clear your doubts about the geometrical configuration.
To facilitate our proof, we added the line segments $DF$to the diagram that portraits the scenario describe in OP’s problem statement. For brevity, we let $\measuredangle ABC=2\beta$, $\measuredangle BCA=2\omega$, and $\measuredangle FKA=\alpha$.
From the properties of the incircle of a triangle, we know that $IB$ and $IC$ are the angle bisectors of $\angle ABC$ and $\angle BCA$ respectively and these lead to,
$$\measuredangle ABI=\measuredangle IBC=\beta \quad\text{and}\tag{1a}$$
$$\measuredangle BCI=\measuredangle ICA=\omega. \tag{1b}$$
We can also deduce that,
$$ID=IE=IF=r, \tag{1c}$$
$$AE=AF,\tag{1d}$$
$$BD-BF, \quad\text{and}\tag{1e}$$
$$CD=CE.\tag{1f}$$
Because of (1a) and (1e), according to SAS congruence rule, the triangle pair $BDK$ and $BKF$ are congruent. So, we have,
$$KD=KF \quad\text{and}\tag{2a}$$
$$\measuredangle BDK=\measuredangle KFB. \tag{2b}$$
Since both angle pairs ($\angle BDK,\space\angle KDP$) and ($\angle KFB,\space\angle AFK$), it follows from (2b) that,
$$\measuredangle KDP=\measuredangle AFK. \tag{3}$$
Furthermore, since $\angle FKA$ and $\angle PKD$ are opposite angles, we shall write,
$$\measuredangle FKA=\measuredangle PKD=\alpha. \tag{4}$$
Using (2b), (3), and (4), and quoting ASA congruence rule, we can state that the triangle pair $FKA$ and $PKD$ are congruent. Hence, we have,
$$AF=DP. \tag{5}$$
It is given that $AE=EC$. Using (1d) and (5), we can infer that,
$$EC=DP.\tag{6}$$
We can use (1f) and (6) to show that, $CP=CQ$, which makes the $\triangle PCQ$ isosceles. Since $CI$ is the angle bisector of the apex angle of the isosceles tringle $PCQ$, it is perpendicular to the base $PQ$ of the said triangle.
Best Answer
This is a synthetic proof using angle chasing mostly. We need to enhance OP’s sketch by introducing point $G$, which is the reflection of the point $A$ about the segment $BC$. We join $G$ to the five points $B$, $C$, $D$, $E$ and $F$ as shown in the diagram. For brevity, let $AB=c$ and $AC=b$, $\measuredangle ABC=\beta$, and $\measuredangle BCA=\omega$. Note that, by construction, $BD=AB=c$ and $CE=AC=b$.
Since $G$ is the reflection of $A$ about $BC$, we have, $$BG=BA=c\qquad\text{and}\qquad CG=CA=b. \tag{1}$$
Furthermore, $$\measuredangle CBG=\measuredangle ABC=\beta, \quad\text{and}\tag{2a}$$ $$\measuredangle GCB=\measuredangle BCA=\omega.\qquad\space \tag{2b}$$
It can be shown that $$\measuredangle BGC =\measuredangle CAB=\pi-\measuredangle ABC-\measuredangle BCA = \pi-\beta-\omega. \tag{3}$$
It is given that $\measuredangle ABD=\dfrac{\pi}{2}$. Therefore, using (2a),we shall write, that $$\measuredangle GBD=\measuredangle ABD-\measuredangle ABC-\measuredangle CAG=\dfrac{\pi}{2}-2\beta. \tag{4a}$$
In similar vein, using (2b), we can obtain, $$\measuredangle ECG =\dfrac{\pi}{2}-2\omega. \tag{4b}$$
According to (1), both triangles $GBD$ and $ECG$ are isosceles. Since we know the apex angles of both of them (see (4a} and (4b)), we can calculate their individual base angles, i.e., $\angle DGB$ and $\angle CGE$, as shown below. $$\measuredangle DGB=\dfrac{\pi-\measuredangle GBD }{2}=\dfrac{\pi}{4}+\beta\tag{5a}$$
Similarly, we can deduce, $$\measuredangle CGE=\dfrac{\pi}{4}+\omega.\tag{5b}$$
Using (3), (5a), and (5b), we can write, $$\measuredangle EGD=\pi - \measuredangle BGC - \measuredangle DGB - \measuredangle CGE =\dfrac{\pi}{2},$$ which means $\triangle EGD $ is a right-angled triangle and, therefore, the center of its circumcircle lies at $F$. This gives us $FG=FD=FE$. Hence, both quadrilaterals $BDFG$ and $CGFE$ are kites. Since the two diagonals of a kite perpendicularly bisect each other, $BF$ and $CF$ are the angle bisectors of $\measuredangle GBD$ and $\measuredangle ECG$ respectively. Therefore, according to (4a) and (4b), we have, $$\measuredangle CBF =\measuredangle CBG +\dfrac{1}{2}\measuredangle GBD = \dfrac{\pi}{4}, \quad\text{and}$$
$$\measuredangle FCB =\measuredangle GCB+\dfrac{1}{2}\measuredangle ECG = \dfrac{\pi}{4}.\qquad\space\space$$ Therefore, $\triangle BFC$ is not only isosceles, but also right-angled.