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Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.

Sagot :

MrRoyal

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

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