Proving angles are equal if bisectors are parallel

The question asks that if there's a parallelogram ABCD and bisectors of < ABC and < ADC are parallel then how can we prove that angles <DAB and <BCD are equal.

I have managed to work backwards and show that triangle DAE and FBC both have the same angles by working with the parallel lines. I have simply sketched in the angles congruent to <DAE and <BCF and gone from there. I let E be the point where the bisector from angle <DAB met the opposite side and F be the point at which the bisectors from <DCB touched the opposite side. How do I prove that it because the proof is dependent on the placement of E and F, That in all cases this will work?

Re: Proving angles are equal if bisectors are parallel

Hello, Idiotinabox!

Is there a typo in the problem?

Quote:

$\displaystyle \text{Given: parallelogram }ABCD\text{ and bisectors of }\angle ABC\text{ and }\angle ADC\text{ are parallel.}$

$\displaystyle \text{Prove that: }\:\angle DAB \:=\: \angle BCD.$

In *any* parallelogram, opposite angles are always equal.

It has nothing to do with angle bisectors.

Code:

` B C`

* - - - - - - *

/ θ /

/ /

/ /

/ /

/ θ /

* - - - - - - *

A D

Re: Proving angles are equal if bisectors are parallel

Problem makes sense if the opposite angle theorem of parallelograms is not allowed

Re: Proving angles are equal if bisectors are parallel

Sorry, it's not a parallelogram. My mistake