From $\angle HEA = 90^\circ = 180^\circ - 90^\circ = 180^\circ - \angle HFA$ (Fig. 5) we deduce that $AEHF$ is cyclic. By definition, $T$ lies on this circle as well, so
$$\angle THE = 180^\circ - \angle TAE = 180^\circ - \angle TAC = \angle ABC,$$
where the final equality is due to the tangent-chord theorem.

Let $K'$ be the intersection of the lines $TH$ and $BC$ (Fig. 6). It's sufficient to show that $K' = K$ or that $K'$ lies on the circumcircles of $TBE$ and $TCF$, as we have $\angle K'HB = \angle THE = \angle ABC$. To show this we notice that
$$\angle ETK' = \angle ETH = \angle EAH = \angle CAH = 90^\circ - \angle ACB = \angle EBC = \angle EBK',$$

Fig. 5

Fig. 6

which shows that the points $T$, $B$, $K'$ and $E$ are concyclic. Analogously we show that $T$, $C$, $K'$ and $F$ are concyclic. Thus $K' = K$, as desired.

Analogously to Solution 1 we show that $\angle THE = \angle ABC$.

Since $\angle BEC = \angle BFC = 90^\circ$, we see that $BCEF$ is cyclic. Thus the lines $TK$, $BE$ and $CF$ are the radical axes of the circumcircles of $BCEF$, $TCF$ and $TBE$. As these intersect at a point, we have that $H$ lies on the line $TK$. This yields $\angle KHB = \angle THE = \angle ABC$, as desired.