A------B----------C----------D------E
C is the midpoint of BD and AE, prove AB is congruent to DE
Since C divides BD and AE, each into two equal halves, we can say
|BC| = |CD|
and
|AC| = |CE|
Now, starting with |AC| = |CE|, we re-write in terms of the shorter segments:
|AB| + |BC| = |CD| + |DE|
Replacing |CD| with |BC| on the RHS:
|AB| + |BC| = |BC| + |DE|
Therefore |AB| = |DE|, and AB is congruent to DE.
