# triangle problem

• Mar 15th 2007, 05:28 AM
balmora55
triangle problem
BF=CF
Angle DFC= Angle AFD
Angle EFB= Angle AFE
Prove that DE is Parallel to CB

• Mar 15th 2007, 10:15 AM
ThePerfectHacker
It is not true.

If the line (which looks like altitute) is not perpendicular as it appears but rather slanated it fails.
• Mar 15th 2007, 11:18 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
It is not true.

If the line (which looks like altitute) is not perpendicular as it appears but rather slanated it fails.

I'm glad someone else noted that. I had thought it to be true but couldn't prove that I wasn't missing something.

-Dan
• Mar 15th 2007, 12:00 PM
balmora55
10x
• Mar 15th 2007, 01:52 PM
ThePerfectHacker
Quote:

Originally Posted by balmora55
sorry to bother you but i have two problems
1. explain what you said in greater detail please
2. i don't know what altitute means
(i'm from israel so i don't know math words in english)

Heir.
• Mar 16th 2007, 01:43 PM
earboth
Quote:

Originally Posted by balmora55
BF=CF
Angle DFC= Angle AFD
Angle EFB= Angle AFE
Prove that DE is Parallel to CB

...

Hello,

I've attached a drawing of a triangle with the given conditions.

(This sign < means angle)

<(EFD) = 90° (it is the half of 180° and therefore constant)

The intersection of AF ∩ DE = P

Now reflect P over FD to P'. Then PP' is parallel to FE because PP' is perpendicular to FD and FD is perpendicular to FE. P' lies on CB because the triangle PFP' is an isosceles triangle. Therefore P'F must be parallel to PE because PEFP' is a parallelogramm.

Do the same procedure with P, P'', F and D.

You get DE is parallel to P'P'' and therefore DE is parallel to BC.

EB