Competition problem — Geometry — Maths Olympiad Prep

Library / /19 of 30

Geometry Difficulty 8.5 Shortlist Prove it Germany

Problem:

Let a triangle $ABC$ be given. A circle $k$ passes through $A$ , intersects the sides $\overline{AB}$ and $\overline{AC}$ again at points $D$ and $E$ respectively, and intersects the side $\overline{BC}$ at points $F$ and $G$ , where $F$ lies between $B$ and $G$ . The tangent to the circle through $B$ , $D$ and $F$ at $F$ and the tangent to the circle through $C$ , $E$ and $G$ at $G$ meet at a point $T$ . We assume that $A \neq T$ holds. Prove that the lines $AT$ and $BC$ are parallel.

Solution

Solution:

By the tangent-chord angle theorem for the chord $\overline{FB}$ we have (see figure) $\angle TFB = 180^{\circ} - \angle BDF = \angle FDA$ . In the cyclic quadrilateral $ADFG$ we therefore have $\angle FDA = 180^{\circ} - \angle AGF = \angle CGA$ (small arcs); analogously $\angle TGB = 180^{\circ} - \angle CGT = \angle GEC$ and $\angle GEC = \angle CFA$ (cyclic quadrilateral $AFGE$ , large arcs). Thus $\angle TFB = \angle CGA$ (1), from which $\angle GFT = \angle AGB$ follows, and $\angle TGB = \angle CFA$ (2).

Thus the triangles $AFG$ and $TFG$ are congruent by ASA, so their altitudes on the common side $\overline{FG}$ are equal, from which the claim follows.

Variant: From (1) and (2) it follows directly that $\angle AFT = \angle AGT$ holds. By the converse of the theorem on cyclic quadrilaterals, this means that $T$ lies on $k$ . With $\angle TAF = \angle TGF = \angle GFA$ the parallelism of $\overline{TA}$ and $\overline{FG}$ follows.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from de; metadata (topic, difficulty) added by this project.