triangle problem

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March 15th 2007, 04:28 AM
balmora55

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triangle problem

BF=CF
Angle DFC= Angle AFD
Angle EFB= Angle AFE
Prove that DE is Parallel to CB

please help
thanks in advance
March 15th 2007, 09:15 AM
ThePerfectHacker

It is not true.

If the line (which looks like altitute) is not perpendicular as it appears but rather slanated it fails.
March 15th 2007, 10:18 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
March 15th 2007, 11:00 AM
balmora55

10x
March 15th 2007, 12:52 PM
ThePerfectHacker

1 Attachment(s)

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.
March 16th 2007, 12:43 PM
earboth

1 Attachment(s)

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