Here are two ways of thinking about this problem. Both
involve what is called proof by contradiction.
In
a
proof
by
contradiction,
you
assume
some
proposition
is
true.
Then,
if
solid,
logical
reasoning
based
on
that
proposition
leads
to
a
contradiction,
that
forces
you
to
admit
that
the
initial
proposition
was
not
true.
Examining each angle: The sum of all three angles in a
triangle is 180o, so the only way for the sum of two of them
to be less than 120o is for the third to be greater than 60o.
IF we could draw a triangle in which the sum of any two
angles is less than 120o, THEN every angle would have
to be greater than 60o (because the sum of the other two
< 120o). This conclusion makes no sense! It’s impossible
for all three angles to be greater than 60o, because their
sum would then be greater than 180o!
This
would
also
mean
that
the
sum
of
any
two
of
them
would
be
greater
than
120o!
This means that we cannot draw such a triangle because
we have seen that IF we could THEN we arrive at a
nonsensical conclusion full of contradictions.
An equivalent algebraic approach: Again, let’s
suppose that we can draw a triangle in which the sum of
any two angles is less than 120o.
We can write down the three conditions that must
simultaneously be true:
1 + 
2 < 120o and
1 + 
3 < 120o and
2 + 
3 < 120o.
Adding all three expressions on the left side, we get:
At the same time, if our conditions (and our initial
assumption) were true, then the sum of the three
expressions on the left should be less than 360o, because
each of them had to be less than 120o. A contradiction!
One way of adding shows that this sum equals 360o; the
other way shows it must be less than 360o. Where can we
have gone wrong?! Only in our initial assumption that all
three conditions can be true simultaneously.
This proves that there is no triangle on the plane in which
a sum of any two angles is less than 120o.
