Question

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

Answer 1

When a line is bisected, the line is divided into equal halves.

See below for the proof of $\mathbf{KE \cong FD}$

The given parameters are:

• AC bisects CD
• IJ bisects CE
• BH bisects ED

By definition of segment bisection, we have:

• $\mathbf{CK \cong KE}$
• $\mathbf{EF \cong FD}$
• $\mathbf{CE \cong ED}$

By definition of congruent segments, the above congruence equations become:

• $\mathbf{CK = KE}$
• $\mathbf{EF = FD}$
• $\mathbf{CE = ED}$

By segment addition postulate, we have:

• $\mathbf{CE = CK + KE}$
• $\mathbf{ED = EF + FD}$

Substitute $\mathbf{ED = EF + FD}$ in $\mathbf{CE = ED}$

$\mathbf{CK + KE = EF + FD}$

Substitute $\mathbf{CK = KE}$ and $\mathbf{EF = FD}$

$\mathbf{KE + KE = FD + FD}$

Simplify

$\mathbf{2KE = 2FD}$

Apply division property of equality

$\mathbf{KE = FD}$

By definition of congruent segments

$\mathbf{KE \cong FD}$

Read more about proofs of congruent segments at:

brainly.com/question/11494126