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l,m,n are three concurrent line concurrent at point A. Given a point B on line l. Is it possible to construct point C on line n such that line m is a median of triangle ABC

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    $\begingroup$ This sounds to me like something that would demand a marked straightedge. But I'm not certain. $\endgroup$ – Arthur Mar 25 '14 at 12:58
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Yes, for sure. Take a point $P$ on $n$: the midpoint of $BP$ is on a line parallel to $n$ through the midpoint of $AB$, so you can find the midpoint of $BC$ by intersecting $m$ with the parallel to $n$ through the midpoint of $AB$.

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There are two cases.In one of the cases it is possible for m to be median.In the other case it is not.

Case 1: When l,m,n are taken in sequence then we can have m as median. By using compass we find two points C, C' on line n which are at same distance from A as distance of B from A.

Case 1

Case 2:While in this case line m becomes an external line. So not possible to find any c on n as median.

Case 2

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