PLEASE HELP!!! :))In isosceles ∆DEK with base DK, EF is the angle bisector of ∠E, m∠DEF = 43°, and DK = 16cm. Find: KF, m∠DEK, m∠EFD.

For a better understanding of the solution given here please find the attached file which has the relevant diagram.To answer this question we will have to make use of the Isosceles Triangle Theorem which states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle. Thus, as a corollary we know that is EF is the angle bisector of the vertex angle ∠E, then, EF is the perpendicular bisector of the of the base DK.Please follow the diagram of a complete understanding of the logic and the solution.As EF is the angle bisector as given in the question, thus we will have:[tex]m\angle DEK=2\times m\angle DE F=2\times 43^{\circ}=86^{\circ}[/tex].Also, from the Theorem we know that KF will be half of DK and thus, KF will be:[tex]KF=\frac{1}{2}DK= \frac{1}{2}\times 16=8[/tex] centimeters.Likewise, from the same theorem we have: [tex]m\angle EFD=90^{\circ}[/tex]