When a line is bisected, the line is divided into equal halves.
See below for the proof of [tex]\mathbf{KE \cong FD}[/tex]
The given parameters are:
- AC bisects CD
- IJ bisects CE
- BH bisects ED
By definition of segment bisection, we have:
- [tex]\mathbf{CK \cong KE}[/tex]
- [tex]\mathbf{EF \cong FD}[/tex]
- [tex]\mathbf{CE \cong ED}[/tex]
By definition of congruent segments, the above congruence equations become:
- [tex]\mathbf{CK = KE}[/tex]
- [tex]\mathbf{EF = FD}[/tex]
- [tex]\mathbf{CE = ED}[/tex]
By segment addition postulate, we have:
- [tex]\mathbf{CE = CK + KE}[/tex]
- [tex]\mathbf{ED = EF + FD}[/tex]
Substitute [tex]\mathbf{ED = EF + FD}[/tex] in [tex]\mathbf{CE = ED}[/tex]
[tex]\mathbf{CK + KE = EF + FD}[/tex]
Substitute [tex]\mathbf{CK = KE}[/tex] and [tex]\mathbf{EF = FD}[/tex]
[tex]\mathbf{KE + KE = FD + FD}[/tex]
Simplify
[tex]\mathbf{2KE = 2FD}[/tex]
Apply division property of equality
[tex]\mathbf{KE = FD}[/tex]
By definition of congruent segments
[tex]\mathbf{KE \cong FD}[/tex]
Read more about proofs of congruent segments at:
https://brainly.com/question/11494126
