What´s the the difference, in degrees, between the measures of the two smallest angles in the triangle below? For reference:  In triangle $ABC$, let $D$ and $E$ be the intersections of the bisectors of angles $ABC$ and $ACB$ with sides $AC$ and $AB$ respectively. Knowing that the measures, in degrees, of angles $BDE$ and $CED$ are equal to $24$ and $18$ respectively, the difference, in degrees, between the measures of the two smallest angles of this triangle is equal to?
My progress:
My drawing

$\angle EID = 180-18-24 = 138^o=\angle CIB \implies \angle DIC = 42^o=\angle BIE\\
\triangle CIB\alpha+\theta+138=180 \therefore \alpha +\theta = 42^o\\
\triangle ABC: \angle A +2\alpha +2\theta  =180^o \implies \angle A= 180 - 2(\alpha+\beta)=180 -2(42) \therefore \angle A = 96^o   $
...???
 A: I believe, there is some trick to simplify solution of the problem with given angles. But I haven't found it.
My solution can be used for any values of given angles and  is based on counting projections of edge FG on direction of edge ED and  direction perpendicular to edge ED.

The point F is point symmetric to E about BD and the point G is point symmetric to D about CE. Then EG = ED = DF, angle FDE is 48°, angle GED is 36°.
To solve the problem we need to find $\alpha$ which is $18°+x$ in my picture. So we need to find $x$. The $\tan x$ can be found as shown in the picture.
For given angles result $x=18°$ requires hard calculations. I was doing these by founding $\sin 18°=\frac{\sqrt{5}-1}{4}$ and using this expression in next calculations.
The final answer is $2\alpha-2\theta=4x-12°=60°$
A: 
Hints: In figure we have:
$\overset{\large\frown}{EG}=24\times2=48^o$
G is midpoint of arc EF so:
$\overset{\large\frown}{EF}=48\times2=96^o$
$\overset{\large\frown}{EF}=\overset{\large\frown}{EA}=\overset{\large\frown}{AK}=2\times 96=192^o$
where K is between D and C on the big circle.
You found $\angle A=96^o$
Using these data you have to show:
$\angle A=\angle E=\angle F=96^o$
Which gives:
$$\angle FCA=360-3\times 96=72$$
so:
$\angle B=(180-96=84)-72= 12^o$
So :
$$\alpha-\theta=36-6=30^o$$
The key point is that CD is perpendicular on EF and $\angle A=\angle E=\angle F=96^o$ and FC is bisector of $\angle E$ which means $\angle AEC=48^o$. In this way:
$\angle AED=30^o|rightarrow \angle ACE=30^o$
which finally gives $\hat{C}=72^o$
Note that EF and EA are tangent on inscribed circle and quadrilateral ACEF is symmetric about EC, this is the reason for $\angle F=\angle A=96^o$.Also BI is perpendicular on EF because the rays of angle $\hat B$ are tangent on inscribed circle and also BI is bisector of that angle.Point F is mirror of point A about CE.
