Menelaus' Theorem and Concurrency

[ebwoolf96]

triangle, parallel

Given \(\displaystyle \triangle ABC\) take \(\displaystyle AD \perp BC\). Consider \(\displaystyle DF \perp AC\), \(\displaystyle DE \perp AB\), \(\displaystyle DM \parallel AB\), \(\displaystyle DN \parallel AC\). Prove that MN, EF, and BC are concurrent.

 

[Deleted member 4993]

Given $\triangle ABC$ take $AD \perp BC$. Consider $DF \perp AC$, $DE \perp AB$, $DM \parallel AB$, $DN \parallel AC$. Prove that MN, EF, and BC are concurrent. View attachment 7822

View attachment 7822
What are your thoughts?

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[ebwoolf96]

View attachment 7822
What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

I am stuck from the beginning!! I extended like MN, FE and BC to a common point P but I am unsure where to go from there.

 

[ebwoolf96]

I am stuck from the beginning. I extended MN, FE and BC to a common point P. However I do not know where to go from there.

 

[ebwoolf96]

Yes

 

[ebwoolf96]

Oh oops. It should Be I'm not sure why it is like that. When I made the diagram it was perpendicular