Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.3 Shortlist Prove it Hong Kong

Let $ABCDEF$ be a convex hexagon such that $\angle ACE = \angle BDF$ and $\angle BCA = \angle EDF$ . Let $A_1 = AC \cap FB$ , $B_1 = BD \cap AC$ , $C_1 = CE \cap BD$ , $D_1 = DF \cap CE$ , $E_1 = EA \cap DF$ and $F_1 = FB \cap EA$ . Suppose $B_1, C_1, D_1, F_1$ lie on the same circle $\Gamma$ . The circumcircles of $\triangle BB_1F_1$ and $\triangle ED_1F_1$ meet at $F_1$ and $P$ . The line $F_1P$ meets $\Gamma$ again at $Q$ . Prove that $B_1D_1$ and $QC_1$ are parallel. (Here, we use $\ell_1 \cap \ell_2$ to denote the intersection point of lines $\ell_1$ and $\ell_2$ .)

Solution

Firstly, since
$\angle EPF_1 = \angle ED_1F_1 = \angle C_1QF_1,$
we have $PE//QC_1$ . By symmetry, we also have $BP//QC_1$ . Therefore, $B, P, E$ are collinear, with $BE//QC_1$ .

Secondly, since $\angle D_1DB_1 = \angle D_1CB_1$ and $\angle EDB = \angle ECB$ , we know that $B_1, C, D, D_1$ and $B, C, D, E$ are two groups of concyclic points. By Reim's theorem, we have $BE//B_1D_1$ (alternatively, since $\angle C_1B_1D_1 = \angle DCE = \angle C_1BE$ ). Thus, $B_1D_1//BE//QC_1$ .

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