BF=CF

Angle DFC= Angle AFD

Angle EFB= Angle AFE

Prove that DE is Parallel to CB

please help

thanks in advance

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- Mar 15th 2007, 04:28 AMbalmora55triangle problem
BF=CF

Angle DFC= Angle AFD

Angle EFB= Angle AFE

Prove that DE is Parallel to CB

please help

thanks in advance - Mar 15th 2007, 09:15 AMThePerfectHacker
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, 10:18 AMtopsquark
- Mar 15th 2007, 11:00 AMbalmora55
10x

- Mar 15th 2007, 12:52 PMThePerfectHacker
- Mar 16th 2007, 12:43 PMearboth
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