# Concurrency Proof

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.

• 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.
– Blue
Oct 9, 2016 at 13:18

## 2 Answers

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.

• I see what you mean....... Thanks. Oct 9, 2016 at 4:05
• Good job! You are welcome! Oct 9, 2016 at 4:06
• There are actually 3 parallelograms.
– Hrhm
Oct 9, 2016 at 4:09
• @Hrhm Yes, exactly, but you need only two of them to prove the concurrency. Oct 9, 2016 at 4:19
• Yes, good point.
– Hrhm
Oct 9, 2016 at 4:24

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