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Constructing triangle $\triangle ABC$ given median to the side $c$ and angles $\alpha$ and $\beta$

I started with the median. Then I constructed a circle to each side of the median, such that the circle is a set of all points which are the vertices of angle $\alpha$ and $\beta$ above the given segment (the median) respectively. Now, if I only could create a line segment $AB$, such that it passes the point $C_1$ (the endpoint of the median and the middle point of side $AB$), each endpoint is on a different circle (to get the correct angles), and the line segments $AC_1$ and $C_1B$ are the same length ($C_1$ is the endpoint of a median), I would be done. Any ideas?

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    $\begingroup$ Consider the midpoint of the centers of the circles. $\endgroup$
    – Blue
    Nov 26 '20 at 9:28
  • $\begingroup$ Got it, thanks! $\endgroup$
    – Jeff
    Nov 26 '20 at 10:20
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So, based on the suggestion of @Blue, the point is to find the midpoint of the centers of the circles $S$, construct a line segment $SC_1$, and then a line perpendicular to $SC_1$ passing through $C_1$. The two intersections with the circles are our points $A$ and $B$.

enter image description here

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    $\begingroup$ +1. That's it! :) ... I'd created an image in case it was needed. I hope you don't mind that I've added it to your answer. (Feel free to remove it.) $\endgroup$
    – Blue
    Nov 26 '20 at 10:23
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    $\begingroup$ That's very kind of you, thank you very much! :) $\endgroup$
    – Jeff
    Nov 26 '20 at 10:34
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    $\begingroup$ Why'd anyone remove your beautiful picture, @Blue ? :) $\endgroup$
    – cosmo5
    Nov 26 '20 at 14:54
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You can also employ this approach: draw a triangle $ABC$ with $\widehat{A}=\alpha$ and $\widehat{B}=\beta$, consider the length of the median $MA$ and apply a homotethy to $ABC$ in such a way that the length of $MA$ becomes the wanted length.

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