Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 4.4 AIME Prove it Croatia

Let $ABCD$ be a square, and let $k$ be the circle centred at $B$ passing through $A$ , $C$ and the point $T$ inside the square. Tangent on $k$ at $T$ intersects the segments $\overline{CD}$ and $\overline{DA}$ at $E$ and $F$ , respectively. Let $G$ and $H$ be the intersections of the lines $BE$ and $BF$ with the segment $\overline{AC}$ , respectively.
Prove that the lines $BT$ , $EH$ and $FG$ are passing through the same point.

Solution

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 $BEF$ .

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