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Geometry Difficulty 8.5 Shortlist Prove it Saudi Arabia

The excircle $\omega_{B}$ of triangle $ABC$ opposite $B$ touches side $AC$ , rays $BA$ and $BC$ at $B_{1}$ , $C_{1}$ and $A_{1}$ , respectively. Point $D$ lies on major arc $\widetilde{A_{1}C_{1}}$ of $\omega_{B}$ . Rays $DA_{1}$ and $C_{1}B_{1}$ meet at $E$ . Lines $AB_{1}$ and $BE$ meet at $F$ . Prove that line $FD$ is tangent to $\omega_{B}$ (at $D$ ).

Solution

Let $F'$ be the intersection point of the tangents to circle $\omega_{B}$ at $B_{1}$ and $D$ .

Let $G$ be the intersection point of lines $B_{1}A_{1}$ and $DC_{1}$ .

Consider the six cyclic points $C_{1}, C_{1}, B_{1}, A_{1}, A_{1}, D$ . Because the tangent lines to $\omega_{B}$ at $C_{1}$ and $A_{1}$ intersect at $B$ , lines $C_{1}B_{1}$ and $A_{1}D$ intersect at $E$ , and lines $B_{1}A_{1}$ and $DC_{1}$ intersect at $G$ , from Pascal theorem, the three points $B$ , $E$ , and $G$ are collinear.

Consider the six cyclic points $C_{1}, B_{1}, B_{1}, A_{1}, D, D$ . Because lines $C_{1}B_{1}$ and $A_{1}D$ intersect at $E$ , the tangent lines to $\omega_{B}$ at $B_{1}$ and $D$ intersect at $F'$ , and lines $B_{1}A_{1}$ and $DC_{1}$ intersect at $G$ , from Pascal theorem, the three points $E$ , $F'$ , and $G$ are collinear.

We deduce that $B$ , $E$ , and $F'$ are collinear and therefore $F' = F$ . This proves that $FD$ is tangent to $\omega_{B}$ at $D$ .

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