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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestStrategy: 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?

Log in to reply

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 :)

Log in to reply

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

Log in to reply