Simple Maths Problem. Help is needed. LC Geomerty

stevemanu

So yea guys dont post here often but really need help with it ! http://www.flickr.com/photos/77493695@N06/9821749446/

Think you have to use similar triangles.
Any help appreciated!

CJC86

To find the perpendicular distance to the line EB, draw a point on it, call it F. Then you will have some similar triangles, as you guessed, since CF will be parallel to both DE and AB. I'll leave the rest to you. Post here again if you get stuck.

stevemanu

Thanks,
Yea Im still stumped... What two triangles should I be looking at in particular to help me get CF ?

sigmundv

stevemanu wrote: »

Thanks,
Yea Im still stumped... What two triangles should I be looking at in particular to help me get CF ?

Triangles ABE and CFE will be similar, so will the triangles DEB and CFE. See if you can come up with some equations based on this.

stevemanu

Ugh this is still bugging me. I have 6/CF = EB/EF 4/CF=EB/FB Too many variables... Need help cracking up here ! Thanks !!

CJC86

Since F is on the line EB, we also have EF + FB = EB. Now you have enough equations to solve for FC.

Tarzann

Use the previous advice to get the ratio of EB to FB.
Then sub this number into 4/CF=EB/FB