Find the angles of given triangle ABC 
A triangle $ABC$ with angle bisectors $AA_1$ and $BB_1$ is given, such that $\angle AA_1B_1 = 24^\circ$ and $\angle BB_1A_1 = 18^\circ$. Find the angles of the triangle.

I've been stuck on this one for quite a long time. After denoting with $I$ the incenter of ABC and deriving that $\angle C = 96^\circ$ from $\angle AIB = 90^\circ + \frac12\angle C = 138^\circ$, I really don't know how to continue. I tried using Geogebra to see everything clearer or at least guess the answer, and I concluded that $\angle A$ and $\angle B$ should be $12^\circ$ and $72^\circ$ respectively, but I'm not sure how to prove it.
Any help would be appreciated. If I come up with something, I will post it right away. Thanks in advance!
 A: Observe: $2(18°)+24° = \color{red}{60°}$. This hints that maybe producing $18°$ might help. Let $A_1$, $B_1$ respectively denote the point of intersection of angle bisectors from A, B to corresponding sides. Since $\angle ABB_1 = \angle CBB_1$, reflect $∆BA_1B_1$ about side $BB_1$. Let the reflection of $A_1$ on AB be $A'$. Due to the property of reflection, $∆BB_1A_1 ≅ ∆BB_1A'$. Let X denote the point of intersection of $A'A_1$ and $BB_1$. $∆A_1XB ≅ ∆A'XB$, so
$\angle A_1XB = \angle A'XB = \frac{180°}{2} = 90°$.
Therefore,
$\angle XA_1A = 90° -(18°+24°) = 48°$
. Observe that $48° = \color{red}{2}×24°$. Since $\angle A_1AB = \angle A_1AC$ reflection $∆A_1AA'$ about line $A_1A$. Let the reflection of $A'$ be $A''$.
$\angle A''A_1A = \angle A'A_1A = 48°$.
Therefore $\angle A''A_1B_1 = 24°$. Let Y denote the point of intersection of $A'B_1$ and $AA_1$. $∆A_1YA' ≅ ∆A_1YA''$. Therefore
$\angle A_1YA'' = \angle A_1YA' = 180°-48°-(90°-18°) = 60°$.
Also $\angle B_1YA = \angle A_1YA' = 60°$. Therefore $\angle B_1YA'' = 180-2×60° = 60°$. Extend $A_1A''$ to a point Z. Observe $B_1$ lies on the angle bisectors of $\angle A''A_1Y$ and $\angle A''YA$. Therefore $A''B_1$ is the angle bisector of $\angle YA''Z$. therefore
$\angle YA''B_1 = 0.5 × (60°+48°) = 54°$.
Therefore
$\angle A_1B_1A'' = 180°-(24°+54°+72°) = 30°$.
$\angle A_1B_1A''$ is exterior angle to $∆A_1B_1A$ therefore $\angle A_1AB_1 = (30°-24°) = 6°$. Therefore $A= 12°$, $B = 72°$, $C=96°$.
A: The following is a kind of reverse approach, beginning with the triangle Euclid uses to construct the regular pentagon.  To avoid confusion, and aid comparison, I keep OP's lettering.
Construct triangle $B_1A_1E$ as the $36^o-72^o-72^o$ triangle of Elements IV, 10.
Let $B_1F$ bisect $\angle EB_1A_1$, and make $\angle B_1A_1I=24^o$. Thus $\angle IB_1A_1=\frac{36}{2}=18^o$.
Next make $\angle B_1ID=\angle DIC=42^o$, and make $IC=IB_1$.
Join and extend $CB_1$ to meet $A_1I$ extended at $A$, and join and extend $AE$ and $CA_1$ to meet at $B$.
Since $\angle AIF=138^o$, and if, as OP notes, $\angle ACB=2(138^o-90^o)=96^o$, while $\angle DCI=180^o-(90^o+42^o)=48^o$, therefore also $\angle ICA_1= 48^o$, $CI$ bisects $\angle ACB$, and the incenter of $\triangle ABC$ lies somewhere on $CI$.
Next, dropping perpendiculars $IG$, $IH$, triangles $HIE$ and $GIA_1$ are congruent. For since $\angle DIA=B_1IA+DIB_1=(18+24)+42^o=84^o$, making $\angle AIE=\angle A_1ID=96^o$, while $\angle GID=84^o$, therefore$$\angle A_1IG=96^o-84^o=12^o$$

In like manner$$\angle HIE=\angle AIE-\angle AIH=96^o-84^o=12^o$$
And since by symmetry $IA_1=IE$, then triangles $HIE$ and $GIA_1$ are congruent and$$IH=IG=ID$$
$I$ is thus the incenter of $\triangle ABC$, and $IA$ bisects $\angle CAB$.
And again, since $\angle DIA=84^o$, and $\angle ADI= 90^o$, therefore$$\angle IAD=6^o$$making$$\angle BAC=12^o$$and$$\angle CBA=72^o$$
