Croatian Junior Mathematical Olympiad

Note that the lines $FA$ and $FT$ are tangent to the circle $k$, hence $|FA|=|FT|$, and the triangles $ABF$ and $TBF$ are congruent. Analogously, $|EC|=|ET|$, and the triangles $CBE$ and $TBE$ are congruent.



Let us denote $\alpha =\angle FBA=\angle FBT$ and $\beta =\angle EBC=\angle EBT$. Since $\angle ABC=90^{\circ }$, it follows that $\alpha +\beta =45^{\circ }$. Now we have $\angle AFB=90^{\circ }-\alpha =45^{\circ }+\beta$ and $\angle AGB=\angle GBC+\angle BCA=\beta +45^{\circ }$, i.e. $\angle AFB=\angle AGB$.

Hence, the quadrilateral $ABGF$ is cyclic, and $\angle BGF=180^{\circ }-\angle AFB=90^{\circ }$, i.e. $FG\perp BE$. Similarly, the quadrilateral $BCEH$ is cyclic, and $EH\perp BF$.

Since the segments $\overline{FG}$ and $\overline{EH}$ are the altitudes of the triangle $BEF$, so as the segment $\overline{BT}$, we finally conclude that the lines $BT$, $EH$ and $FG$ are passing through the same point - the orthocentre of the triangle $BEF$.