Estonian Mathematical Olympiad

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 $K^{\prime }$ be the intersection of the lines $TH$ and $BC$ (Fig. 6). It's sufficient to show that $K^{\prime }=K$ or that $K^{\prime }$ lies on the circumcircles of $TBE$ and $TCF$, as we have $\angle K^{\prime }HB=\angle THE=\angle ABC$. To show this we notice that $$\angle ET{K}^{\prime }=\angle ETH=\angle EAH=\angle CAH=90^{\circ }-\angle ACB=\angle EBC=\angle EB{K}^{\prime },$$ Fig. 5 Fig. 6

which shows that the points $T$, $B$, $K^{\prime }$ and $E$ are concyclic. Analogously we show that $T$, $C$, $K^{\prime }$ and $F$ are concyclic. Thus $K^{\prime }=K$, as desired.

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Alternative solution.

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.