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In a rhomboid $ABCD$ an angle bisector is $DM$ ($M\in BC)$. If $AB=6$, compute the length of the segment that joints the midpoints of $AM$ and $BD$

I don't know how to proceed in this kind of problem, i never worked with the angle bisector in a rhomboid. I tried to use the fact that sum of contiguous angles is $180º$, named the length of all the segments with variables and used the fact that the diagonals bisects each other, but it didn't work.

Any hints?

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  • $\begingroup$ If $DM$ is an angle bisector so $M\equiv B$. $\endgroup$ Jul 7 '19 at 5:33
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If $DM$ is an angle bisector so $M\equiv B$, which gives the answer: $$\frac{6}{2}=3.$$

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