First, we will show that the line $PJ$ passes the midpoint of a side $AB$.
Let $M$ be the intersection of two lines $PJ$ and $AB$, and $S$ be the intersection of two lines $BP$ and $AI$. By using Menelaus' theorem to the triangle $ABS$ and the line $PJ$, we have
$$\frac{AM}{BM} \cdot \frac{BP}{PS} \cdot \frac{SJ}{JA} = 1. \qquad (1)$$
Also, by the angle bisector theorem, we get
$$\frac{SJ}{JA} = \frac{BS}{AB}. \qquad (2)$$
In other hand, since $\angle JBD = \frac{\angle C}{2}$ and $\angle DEI = \angle DCI = \frac{\angle C}{2}$, four points $J, B, E, D$ are cyclic, therefore $\angle BEJ = \angle BDJ = \frac{\angle B}{2}$. From the fact $AB \parallel PE$, we have
$$\angle BEP = \angle BEJ + \angle PEA = \angle BEJ + \angle EAB = 90^\circ - \frac{\angle C}{2}.$$
By the fact $AB \parallel PE$ again, we know $\angle EPB = \angle PBA = \angle C$, hence we get
$$\angle PBE = 180^\circ - \angle C - \left(90^\circ - \frac{\angle C}{2}\right) = 90^\circ - \frac{\angle C}{2},$$
so we obtain $\angle PBE = 90^\circ - \frac{\angle C}{2} = \angle PEB$ and therefore
$$PB = PE. \qquad (3)$$

The fact $AB \parallel PE$ also implies
$$\frac{BS}{AB} = \frac{PS}{PE}. \qquad (4)$$
From (1), (2), (3), (4), we get $AM = BM$, and $M$ is the midpoint of the line segment $AB$, as desired.
Now let $N$ be the midpoint of the side $AC$. Then since $ABC$ is similar to $ADB$, we have $\angle AJM = \angle AIN$, and by the midpoint theorem we get $\angle AIN = \angle AA'Q$. We already know that the points $M, J, P, Q$ are collinear, so $\angle QJA' = \angle AJM$ holds, and we get $QJ = QA'$.