Part 3
we know that
If EF is a bisector
then
mand
msubstitute
msubstitute the given values
5x+19=2(3x+1)
solve for x
5x+19=6x+2
6x-5x=19-2
x=17
Part 4
we have that
If EF is a bisector
then
msubstitute the given values
5x-3=2x+15
solve for x
5x-2x=15+3
3x=18
x=6