Proving similar triangles within a trapezoid ABCS is a trapezoid, and EQ and FP are bisectors of these trapezoid arms.
I would like to prove that angle CQB is equal to angle APD.

I tried to show similarity between triangles CQB and APD, because both of them are isosceles. If I could show that similarity exists, then angle CQB could equal to angle APD. However, I don't know how to show that these triangles are similar.
 A: You have to prove that minor base of trapezoid and vertices of isosceles triangles are on the same circle and major base and vertices of isosceles triangles are on another circle. Thus the angles in green are equal and the angle in pink are equal so the sum of green and pink are equal and finally the two angles of the isosceles triangles are equal

A: List of things to be done:-

1) Is FE // AB // DC? If yes, then $\theta = \theta’$.
2) Is FQPE cyclic? If yes, then $\alpha = \alpha’$ and $\beta = \beta’$.
3) What is the relation between $\theta’$, $\alpha$, and $\beta$?
4) Summing up the above, is DQPC cyclic? That is, will $\alpha’ + \beta’ = \theta$?.
5) If it is a yes to the above step, then x = y. The required result follows.
A: The other two answers leave a lot of steps missing, in particular showing the PEFQ is a cyclic quadrilateral, so I'll focus on that part.

This part does not require a trapezium. Just take two line construct perpendiculars at point E and F, find the points where these intersect the other line P and Q and also the point where the perpendiculars intersect R.
First, consider triangles FRQ and ERP. Both have a right angle and the same angle at R, hence they are similar triangles. As they are similar the corresponding lengths are multiples of each other. So if FR and QR are of lengths l and m lengths ER and PR are λl and λm. 
Next, consider FRE and QRP. Again these have the same angles at R. As the ratio of side lengths are equal the two triangles are also similar.
Now we can draw on the different angles. 

Considering the quadrilateral EFQP for it to be cyclic we require the sum of opposite angles to be 180°. Indeed
EFQ + QPE = (β + 90°) + (α + γ) = α + β + γ + 90°

PEF + FQP = (γ + 90°) + (α + β) = α + β + γ + 90°

So both pairs of opposites angles are equal and must be 180°. Hence we have a cyclic quadrilateral.
The next step is to consider the trapezium ABCD. As F and E are both midpoints, FE is parallel to AB and also to CD. Angles CDA = EFQ and BCD = PEQ so CDQP is also cyclic. It also easy to show ABPQ is cyclic. Then we can proceed like Raffaele to the final result.
