Solution?

Let all tangents at $K, L, M$ intersect on $X$. From the construction it is obvious that $X, A, B$ points are collinear and $XK=XL$. From the given condition $KB=BL$. Combining this two gives us $XKBL$ is a deltoid and so $XB \perp KL$. By angle chasing $\angle BAL= \angle KAB = \angle KLB$. Let $KL \cap AB = \{P\}$. Then we have $\triangle BAL \sim \triangle BLP$. Since $\angle LAB = 90$, $\angle ALB = 90$ which implies $[AB]$ is diameter.

Let $T$ be the intersection of tangents to circles $\omega_1$ and $\omega_2$ from points $K$ and $L$ respectively. Obviously, the line $AB$ passes through $T$. Note that $\angle BKL=\angle BAL$, $\angle BKT=\angle KAB$, meaning BK is the angle bisector of $\angle TKL$. By the same way we get that $LB$ is the angle bisector of $\angle KLT$. So $B$ is the incenter of triangle $TKL$, meaning $TB$ is the angle bisector of $\angle KTL$. However the center $O$ of $\omega_3$ lies on the angle bisector of $\angle KTL$, hence $A$, $O$, $B$ are collinear. This is the same as saying that $AB$ is the diameter of $\omega_3$

By Archimedes lemma: KA/KB=MA/MB=LA/LB, KB=LB => KA=LA, so triangles are equal and <ALB=<AKB=180-<ALB => <ALB=90 (sorry, i don’t know LateX)