CA and DB are diameters of the circle and EC, ED, EB, and EA are radii which means EC=ED=EB=EA which means they are isosceles triangles and since angle CED is congruent to angle AEB we have triangle AEB is congruent to triangle CED.
Now we need ti find z.
Let's focus on triangle CED. Since triangle CED is an isosceles triangle, angle C and D are congruent which means
16z+12+2z+4+2z+4 = 180
20z+20=180
20z=160
z=8
Then solve for angle D
2z+4 = D
16+4 = D
D = 20
And since triangles CED and AEB are congruent Angle D = Angle A so Angle A = 20
:)))))))))))))))))))))
Wait oops you asked for mBA
If thats the case it is quite simple. We already know z = 8 and angle CED = angle AEB and since CED is 140 that means AEB is 140
SO IN CONCLUSION
mBA = 140
Wow i wrote alote
(2z+4)*2 + 16z+12 = 180 (inscribed angles intercept arcs twice their measure)
z = 8
Arc BA = 16z+12 = 140 degrees