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NMO Selection Tests for JBMO

Geometry Difficulty 6.4 National Olympiad Prove it Romania

Let $ABC$ be a triangle with $AB < AC$ and $\omega$ be its circumcircle. The tangent at $A$ to the circle $\omega$ intersects line $BC$ at $D$ , and the line through $B$ parallel to $AD$ meets again the circle $\omega$ at $E$ . Line $DE$ intersects $AB$ at $F$ and the circle $\omega$ for the second time at $G$ . On the line $BE$ , consider the point $N$ such that $B, G, F$ , and $N$ are con-cyclic. Let $S$ and $T$ be the intersection points of line $FN$ with $AD$ and $AE$ , respectively. Prove that $STDG$ is a cyclic quadrilateral.
JBMO Shortlist

Solution

$\angle TAB = \angle BCE = 180^\{\circ\} - \angle BGE = 180^\{\circ\} - \angle BGF = \angle BNF = \angle BNT,$
so ANBT is also cyclic.

Using the fact that $AD \parallel BE$ , it follows that $\angle ATS = \angle ABN = \angle BAD = \angle FAS$ , so line AB is the tangent at A to the circumcircle of AST, (1).

Also, from $AD \parallel BE$ it results that $\triangle FAD \sim \triangle FBE$ , so $\frac{FA}{FB} = \frac{FD}{FE}$ . By power of point F to circle $\omega$ , we infer that $FA \cdot FB = FE \cdot FG$ . Multiplying the last two equalities, we get $FA^2 = FD \cdot FG$ .

From (1) it follows that $FS \cdot FT = FA^2 = FD \cdot FG$ , so S, T, D, G belong to the same circle. Hence, STDG is a cyclic quadrilateral.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.