In the diagram, chords DE, EF and DF are drawn in the circle with centre O - NSC Mathematics - Question 11 - 2021 - Paper 2

To prove that DFK = Ē, we first note that angle BFK equals 90° because it's a tangent at point F.

By the property of circles, we also know that angle DFB = angle BFD. Thus, we can set up the equation:

$DFK = 90° - BFD$

Since BFD is opposite to EF which is subtended by angle DE, we can substitute to find:

$DFK = 90° - (angle subtended by DE)$.

Since both angles subtended by the same chord are equal in a circle, it follows that DFK = Ē.