We're given LM = NO which will be used in substitution later.
By the segment addition postulate, we can write
LN = LM+MN
which basically says "glue LM and MN together to get LN". All three segments fall on the same line.
Now substitute or replace LM with NO. This works because LM = NO is given
So we go from this
LN = LM+MN
to this
LN = NO+MN
Rearrange terms to go from
LN = NO+MN
to
LN = MN+NO
The formal property used is the "Commutative Property of Addition"
Now notice on the right hand side we can combine MN and NO to get MO. Again this is using the segment addition postulate.
So the last step is going from
LN = MN+NO
to
LN = MO
Have a look at the attached image to see how to format this proof into a two-column proof.
By the segment addition postulate, we can write
LN = LM+MN
which basically says "glue LM and MN together to get LN". All three segments fall on the same line.
Now substitute or replace LM with NO. This works because LM = NO is given
So we go from this
LN = LM+MN
to this
LN = NO+MN
Rearrange terms to go from
LN = NO+MN
to
LN = MN+NO
The formal property used is the "Commutative Property of Addition"
Now notice on the right hand side we can combine MN and NO to get MO. Again this is using the segment addition postulate.
So the last step is going from
LN = MN+NO
to
LN = MO
Have a look at the attached image to see how to format this proof into a two-column proof.
