# Unsure of how to entitle this problem.

• May 24th 2011, 03:41 PM
Bashyboy
Unsure of how to entitle this problem.
Okay, I have been ruminating over this problem for quite a while now and, to be frank, I am considerably exasperated. (Okay maybe that is a slight exaggeration). But I have decided to stop scratching my head in bewilderment and seek some help. Okay the problem goes as follows:

"Two beams, AB and CD, are parallel. Find the distance AE"

Problem2.png picture by Bashyboy - Photobucket

I couldn't upload the picture, it wouldn't let me, so hopefully this will link will work. I have not undergone any formal studying in the field of Geometry, so--if it's not too much to ask--could someone please give me detailed steps into this that would be greatly appreciated.
• May 24th 2011, 04:13 PM
TheEmptySet
Quote:

Originally Posted by Bashyboy
Okay, I have been ruminating over this problem for quite a while now and, to be frank, I am considerably exasperated. (Okay maybe that is a slight exaggeration). But I have decided to stop scratching my head in bewilderment and seek some help. Okay the problem goes as follows:

"Two beams, AB and CD, are parallel. Find the distance AE"

Problem2.png picture by Bashyboy - Photobucket

I couldn't upload the picture, it wouldn't let me, so hopefully this will link will work. I have not undergone any formal studying in the field of Geometry, so--if it's not too much to ask--could someone please give me detailed steps into this that would be greatly appreciated.

Since the two lines are parallel using alternate interior angles we know that

\$\displaystyle <EAB = <EDC\$

It is the same for the opposite angles as well.

This states that the two triangles are similar. Use this to find the length of the Segment AE.

After this form the right triangle DAB and use the Pythagorean theorem to find the distance you want.
• May 25th 2011, 05:47 AM
Bashyboy
Thanks, I appreciate it.
• May 25th 2011, 05:53 AM
Ackbeet
Quote:

Originally Posted by TheEmptySet
...\$\displaystyle <EAB = <EDC\$...

Note that

[TEX]\angle EAB=\angle EDC[/TEX] yields

\$\displaystyle \angle EAB=\angle EDC.\$
• May 25th 2011, 06:10 AM
TheEmptySet
Awesome thanks! I will add that to my LaTex vocabulary!
• May 25th 2011, 06:18 AM
Ackbeet
You're welcome!