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Prove that the line joining the midpoints of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals are concurrent.

I haven't really done concurrency proofs before, but I guess the first part is because the midpoints of sides of any quadrilateral form a parallelogram and diagonals of a parallelogram are concurrent. I need some help with the second part.

Addendum- In any triangle $ABC$, prove that the bisectors of the interior angle $A$ and exterior angles at $B$ and $C$ are concurrent.

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  • $\begingroup$ Since the Addendum has nothing to do with the original question, it "should" be posted separately; however, since you've already provided an answer, it doesn't matter. I'll note that the Addendum follows immediately from the trigonometric form of Ceva's Theorem. $\endgroup$
    – Blue
    Oct 9, 2016 at 13:18

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Hint. Indeed it's all about two different parallelograms formed by two groups of four of these six midpoints. And it is true that the diagonals of a parallelogram intersect, but their intersection point is very special. Special for both diagonals.

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  • $\begingroup$ I see what you mean....... Thanks. $\endgroup$ Oct 9, 2016 at 4:05
  • $\begingroup$ Good job! You are welcome! $\endgroup$ Oct 9, 2016 at 4:06
  • $\begingroup$ There are actually 3 parallelograms. $\endgroup$
    – Hrhm
    Oct 9, 2016 at 4:09
  • $\begingroup$ @Hrhm Yes, exactly, but you need only two of them to prove the concurrency. $\endgroup$ Oct 9, 2016 at 4:19
  • $\begingroup$ Yes, good point. $\endgroup$
    – Hrhm
    Oct 9, 2016 at 4:24
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The answer to the addendum of the question is actually quite beautiful. Those who want to see the answer can go here-
http://www.artofproblemsolving.com/community/c2578

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