A Geometry Proof Problem

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

\( \textbf{Problem:} \)

Let M M be an arbitrary point lying outside of parallelogram ABCD ABCD . Consider line l1 l_1 passing through A A parallel to MC MC , line l2 l_2 passing through B B parallel to MD MD , line l3 l_3 passing through C C parallel to MA MA and line l4 l_4 passing through D D and parallel to MB MB . Prove that all 4 4 lines l1,l2,l3,l4 l_1, l_2, l_3, l_4 share a common point.

Thoughts: \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, 9 months ago

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Strategy: Define l1l2=Nl_1\cap l_2=N, prove that NN lies on the other two lines. btw: there's a form of symmetry happening here, what is it?

Xuming Liang - 5 years, 9 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 N . Thank you for the hint :)

Matthew Kendall - 5 years, 9 months ago

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

Hjalmar Orellana Soto - 5 years, 9 months ago

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