# Given four points $A, B, C, D$ in a straight line, find a point $O$ in the same straight line such that $OA: OB = OC:OD$

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.

• First, given a point P in (AB), how would you construct Q in (AB) such that PA:PB=QC:QD ? – Joce Apr 15 at 6:33
• 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. – Mark Bennet Apr 15 at 7:16

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 $$AA'$$).