Hello, Idiotinabox!
Is there a typo in the problem?
In any parallelogram, opposite angles are always equal.
It has nothing to do with angle bisectors.
Code:B C * - - - - - - * / θ / / / / / / / / θ / * - - - - - - * A D
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?
Hello, Idiotinabox!
Is there a typo in the problem?
In any parallelogram, opposite angles are always equal.
It has nothing to do with angle bisectors.
Code:B C * - - - - - - * / θ / / / / / / / / θ / * - - - - - - * A D