Video Transcript
𝐴𝐵𝐶 is a triangle where 𝐹 is
the midpoint of line segment 𝐵𝐶. Is 𝐴𝐵 plus 𝐴𝐶 less than, equal
to, or greater than 𝐴𝐹 plus 𝐵𝐹?
Let us start off by sketching a
diagram with an arbitrary triangle 𝐴𝐵𝐶. We will also draw the midpoint 𝐹
of the line segment 𝐵𝐶. And let us also draw the line
segment 𝐴𝐹. Now, the question is asking us to
compare 𝐴𝐵 plus 𝐴𝐶 to 𝐴𝐹 plus 𝐵𝐹.
One theorem that will be of great
use to us here is the triangle inequality, which states that any side length in a
triangle is shorter than the sum of the two remaining lengths. In this example, we can apply this
inequality to say that 𝐵𝐶 is less than 𝐴𝐵 plus 𝐴𝐶 because they are the three
side lengths of the triangle.
Notice that the right-hand side of
this inequality is the same expression as in the question. So, the question is, can we link
𝐵𝐶 to 𝐴𝐹 plus 𝐵𝐹 in any way? Well, we know that 𝐵𝐶 is equal to
𝐵𝐹 plus 𝐹𝐶 since they are the two halves of this line segment. And since the two line segments are
congruent, this is equal to two 𝐵𝐹.
Now, in order to proceed with this
line of logic, we need to consider two separate cases. Let us suppose that 𝐴𝐹 is less
than or equal to 𝐵𝐹. Then, by adding 𝐵𝐹 to both sides,
this assumption is equivalent to 𝐴𝐹 plus 𝐵𝐹 is less than or equal to two
𝐵𝐹. Subsequently, by using the
inequality we have already calculated, we have that 𝐴𝐹 plus 𝐵𝐹 is strictly less
than 𝐴𝐵 plus 𝐴𝐶. And if we reverse this inequality,
we have shown that 𝐴𝐵 plus 𝐴𝐶 is greater than 𝐴𝐹 plus 𝐵𝐹. So, we have solved this question
for one case, but we also need to consider the case where 𝐴𝐹 is greater than
𝐵𝐹.
Let us clear some space to consider
this scenario. Let us first draw a triangle where
𝐴𝐹 is greater than 𝐵𝐹. Now, suppose that we form a
parallelogram, 𝐴𝐵𝐴 prime 𝐶. So, the opposite sides 𝐴𝐶 and
𝐵𝐴 prime are congruent, and similarly, 𝐴𝐵 and 𝐶𝐴 prime are congruent. We can also extend 𝐴𝐹 onwards to
get 𝐴𝐴 prime. Now, since 𝐴𝐶𝐴 prime is a
triangle, by the triangle inequality, we have that 𝐴𝐴 prime is less than 𝐴𝐶 plus
𝐶𝐴 prime. Also, since this is a
parallelogram, 𝐶𝐴 prime is equal to 𝐴𝐵.
We can also use a similar trick to
before and rewrite 𝐴𝐴 prime as 𝐴𝐹 plus 𝐹𝐴 prime. And since 𝐹 is the midpoint of
this line, this is equal to two 𝐴𝐹. Finally, we can use the fact that,
in this case, 𝐴𝐹 is greater than 𝐵𝐹. Thus, we have that 𝐴𝐹 plus 𝐵𝐹
is less than two 𝐴𝐹, which is less than 𝐴𝐶 plus 𝐴𝐵.
So, in conclusion, by combining the
left- and right-hand sides of the above inequality, we have shown in all cases that
𝐴𝐵 plus 𝐴𝐶 is greater than 𝐴𝐹 plus 𝐵𝐹.
