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I've been on this problem for an entire day. The only thing I learnt is that $O$ is between $A$ and $B$, I feel like I am missing something simple.

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  • $\begingroup$ First, given a point P in (AB), how would you construct Q in (AB) such that PA:PB=QC:QD ? $\endgroup$ – Joce Apr 15 at 6:33
  • $\begingroup$ One way of making progress (not necessarily the best or the easiest) is to introduce co-ordinates on the line, with $x$ the co-ordinate of $O$ you have $\frac {a-x}{b-x}=\frac {c-x}{d-x}$ which reduces to a linear equation [ignoring the infinite solution] and identifies the point you are looking for. $\endgroup$ – Mark Bennet Apr 15 at 7:16
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Make a parallel copy of the line, containing points $A', B',C', D'$. Let $Q$ be the intersection of $A'C$ and $B'D$. Then for any line $\ell $through $Q$, let $O'$ and $O$ be the intersection points with the two parallel lines. Then $O'A':O'B'=OC:OD$. Now what you want is that $O'$ is the parallel copy of $O$, i.e., take $\ell$ perpendicular to the given line (i.e., parallel to $AA'$).

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  • $\begingroup$ Thanks! Nice trick. Can't believe I missed that. $\endgroup$ – Raj_27 Apr 15 at 7:09

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