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This is part of a bigger problem. We are given three points $A,B,C$ in general position. The goal is to construct a triangle $AEF$ such that the segment $AB$ is a median and $C$ is the orthogonal projection of $E$ on $AF$.

We trace the line $AC$, we know $F$ will lie on this line. We can then trace the altitude by constructing the perpendicular to $AC$ passing through $C$, we know $E$ will lie on this new line.

But I am now stuck with two lines and a point $B$. I need to find a segment whose endpoints lie on each line and whose middle is $B$. I have tried for quite a while and am not sure if this is possible.

Does anybody has an idea? Thanks in advance!

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H. Potter
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    $E$ is on the perpendicular of $AC$ and $F$ is on the line $AC$ then we just want $E$ , $F$ such that $|EB| = |BF|$ right ? because $AB$ is a median. – IrbidMath Mar 03 '20 at 15:14

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Hint: median to hypotenuse of a right triangle is half of hypotenuse. Consider right triangle $CEF$ and median $CB$. If you can find $CB$ then $EF=2CB$.

Vasili
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  • I was about to comment: "That means that, in a right triangle, the middle of the hypotenuse is the center of the circumscribed circle, that's crazy". It's at this moment I realized that it was just a disguised version of Thales' theorem. Thanks! – H. Potter Mar 03 '20 at 15:38
  • @H.Potter: Yest, it's also the center of the circumscribed circle. Pretty amazing result! – Vasili Mar 03 '20 at 15:49