The measure.

${\text{Since, AB = AD}} \implies \angle ABD = \angle ADB \quad ................(1) \\ In \triangle ABD, \\ \angle BAD + \angle ABD + \angle BDA = 180 \\ 44 + \angle ABD + \angle BDA = 180 \\ \angle ABD + \angle BDA = 136 \\ So, \angle ABD = \angle ADB = \dfrac{136}{2} = 68 \quad (By (1)) \\ {\text{In }} \triangle BDC, {\text{we have}} \\ ext\angle ADB = \angle DBC + \angle DCB \\ \therefore \angle DBC = \color{#3D99F6}{\boxed{45}}$

DBC=(180-A)/2-C -> DBC=45

Triangle $ABD$ is isoceles, so $\angle ABD = \angle ADB = (180 ^\circ - 44 ^\circ)/2 = 68 ^\circ$ .

Knowing that two angles on the straight line $AC$ sum to $180 ^\circ$ :

$\angle BDC = 180 ^\circ - \angle ADB = 180 ^\circ - 68 ^\circ = 112 ^\circ$ .

Now we solve for the solution, $\angle DBC$ , knowing the angles in the triangle $BDC$ sum to $180 ^\circ$ :

$\angle DBC = 180 ^\circ - \angle BDC - \angle DCB = 180 ^\circ - 112 ^\circ - 23 ^\circ = 45 ^\circ$ .