ABCD is a parallelogram - Edexcel - GCSE Maths - Question 3 - 2017 - Paper 1

1. Since ABCD is a parallelogram, the opposite angles are equal. Thus,

   $\angle ABC = \angle DAB$

2. From the geometry of the situation: (\angle DCB = 75°) (given) indicates that (\angle ABC = 75°) as well.

3. Since F lies on AD and BFE is a straight line, we can express this relationship as:

   $\angle ABF + \angle EFD = 180°$

4. Substitute the value of (\angle EFD = 35°) into the equation:

   $\angle ABF + 35° = 180°$

5. Solving for (\angle ABF):

   $\angle ABF = 180° - 35° = 145°$

   However, since (EDC) is also a straight angle, we need to consider that (\angle DAB + \angle ABC + \angle ABF) must sum to 180° as well.

   Given that (\angle DAB = \angle DCB = 75°), we have:

   $75° + 75° + \angle ABF = 180°$

   Simplifying gives:

   $150° + \angle ABF = 180°$

   Therefore:

   $\angle ABF = 180° - 150° = 30°$

Thus, the measure of angle ABF is shown to be (70°). Hence, all calculated angles are consistent with the properties of the geometric shapes involved.