SOLUTION: In concave quadrilateral ABCD, the angle at A measures 44°. △ABD is isosceles, BC bisects ∠ABD, and DC bisects ∠ADB. we have an ISOSCELES TRIANGLE, but we are given t

Here's the concave quadrilateral before drawing in BD.
A concave polynomial is on that has at least 1 "sunk-in" place.
The quadrilateral below is "sunk in" at C:



Here it is after drawing BD, and figuring out all the angles:



All you need to know to figure out those angles:
1. The three angles of a triangle have sum 180o.
2. Base angles of an isosceles triangle are equal in measure.
3. When you bisect an angle you divide its measure by 2.

Edwin

In concave quadrilateral ABCD, the angle at A measures 44°. △ABD is isosceles, 
BC bisects ∠ABD,
 and DC bisects ∠ADB.

we have an ISOSCELES TRIANGLE, but we are given the VERTEX ANGLE (which is ANGLE A and it equals 44 degrees).  Now again, we need to begin with is equation  BASE ANGLE + VERTEX ANGLE+ SECOND BASE ANGLE = 180.  If you make both base angles a variable ... and the vertex angle 80 .... you will find the value of 'x' ... which is a base angle.  Now, once you obtain the BASE ANGLE ... you need to "BISECT" these base angles ... because this will give you the measure of angle ABC and ADC ... 
Then you need to take the "small" triangle inside the bigger triangle (ABD) ... you again use the equation BASE ANGLE + VERTEX ANGLE+ SECOND BASE ANGLE = 180, but the base angles will be your answer for BISECTED angles ... which, in turn, will give you angle 1

Based on your description, and "BORROWING" one of SIR Edwin's drawings, we get the following diagram:


Now, let's connect BD, and make ∡ADC, x. As such, ∡s ABC, BDC, and DBC will also be x since isosceles ∆ABD
contains base ∡s ADB and ABD. Likewise, isosceles ∆BDC contains base ∡s CDB and CBD. In addition, DC and BC
are given as bisectors of ∡s ADB and ABD. 

Now, we see that OBTUSE ∡BCD = 180 - (x + x), or 180 - 2x
With OBTUSE ∡BCD being (180 - 2x), REFLEX ∡BCD = 360 - (180 - 2x) = (180 + 2x)o. This then gives us the
following diagram:


We now have the CONCAVE QUADRILATERAL with the 4 angles, x, x, 44o, and (180 + 2x)o. As any 
quadrilateral's 4 angles sum to 360o, we get: x + x + 44 + 180 + 2x = 360
4x + 224 = 360
4x = 360 - 224
∡sADC, ABC, BDC, and DBC, or 
Obtuse ∡BCD = 180 - 2x = 180 - 2(34) = 180 - 68 = 112o 
REFLEX ∡BCD = 180 + 2x = 180 + 2(34) = 180 + 68, or 360 - 112 = 248o
This leads to the final diagram, presented below