# A Geometry Proof Problem

Hi, I have been stuck on this problem, please help me.

$$\textbf{Problem:}$$

Let $M$ be an arbitrary point lying outside of parallelogram $ABCD$. Consider line $l_1$ passing through $A$ parallel to $MC$, line $l_2$ passing through $B$ parallel to $MD$, line $l_3$ passing through $C$ parallel to $MA$ and line $l_4$ passing through $D$ and parallel to $MB$. Prove that all $4$ lines $l_1, l_2, l_3, l_4$ share a common point.

$\textbf{Thoughts:}$

I am lost in how to really complete the proof of this problem, but I did notice that there are many more parallelograms in this picture. I'm not sure if this is helpful in any way, but it would be awesome if anyone has any other ideas. Note by Matthew Kendall
5 years, 8 months ago

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## Comments

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There are a lot of parallel lines, you should try by looking for angles

- 5 years, 8 months ago

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Strategy: Define $l_1\cap l_2=N$, prove that $N$ lies on the other two lines. btw: there's a form of symmetry happening here, what is it?

- 5 years, 8 months ago

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I found it, there is a reflection of two congruent triangles through the center of the parallelogram. Once you prove that, the other two lines definitely would intersect at $N$. Thank you for the hint :)

- 5 years, 8 months ago

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