# Question Video: Solving Inequalities Using the Triangle Inequality Theorem Mathematics

π΄π΅πΆ is a triangle where πΉ is the midpoint of line segment π΅πΆ. Is π΄π΅ + π΄πΆ <, =, or > π΄πΉ + π΅πΉ?

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### Video Transcript

π΄π΅πΆ is a triangle where πΉ is the midpoint of line segment π΅πΆ. Is π΄π΅ plus π΄πΆ less than, equal to, or greater than π΄πΉ plus π΅πΉ?

Let us start off by sketching a diagram with an arbitrary triangle π΄π΅πΆ. We will also draw the midpoint πΉ of the line segment π΅πΆ. And let us also draw the line segment π΄πΉ. Now, the question is asking us to compare π΄π΅ plus π΄πΆ to π΄πΉ plus π΅πΉ.

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 π΅πΆ is less than π΄π΅ plus π΄πΆ 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 π΅πΆ to π΄πΉ plus π΅πΉ in any way? Well, we know that π΅πΆ is equal to π΅πΉ plus πΉπΆ since they are the two halves of this line segment. And since the two line segments are congruent, this is equal to two π΅πΉ.

Now, in order to proceed with this line of logic, we need to consider two separate cases. Let us suppose that π΄πΉ is less than or equal to π΅πΉ. Then, by adding π΅πΉ to both sides, this assumption is equivalent to π΄πΉ plus π΅πΉ is less than or equal to two π΅πΉ. Subsequently, by using the inequality we have already calculated, we have that π΄πΉ plus π΅πΉ is strictly less than π΄π΅ plus π΄πΆ. And if we reverse this inequality, we have shown that π΄π΅ plus π΄πΆ is greater than π΄πΉ plus π΅πΉ. So, we have solved this question for one case, but we also need to consider the case where π΄πΉ is greater than π΅πΉ.

Let us clear some space to consider this scenario. Let us first draw a triangle where π΄πΉ is greater than π΅πΉ. Now, suppose that we form a parallelogram, π΄π΅π΄ prime πΆ. So, the opposite sides π΄πΆ and π΅π΄ prime are congruent, and similarly, π΄π΅ and πΆπ΄ prime are congruent. We can also extend π΄πΉ onwards to get π΄π΄ prime. Now, since π΄πΆπ΄ prime is a triangle, by the triangle inequality, we have that π΄π΄ prime is less than π΄πΆ plus πΆπ΄ prime. Also, since this is a parallelogram, πΆπ΄ prime is equal to π΄π΅.

We can also use a similar trick to before and rewrite π΄π΄ prime as π΄πΉ plus πΉπ΄ prime. And since πΉ is the midpoint of this line, this is equal to two π΄πΉ. Finally, we can use the fact that, in this case, π΄πΉ is greater than π΅πΉ. Thus, we have that π΄πΉ plus π΅πΉ is less than two π΄πΉ, which is less than π΄πΆ plus π΄π΅.

So, in conclusion, by combining the left- and right-hand sides of the above inequality, we have shown in all cases that π΄π΅ plus π΄πΆ is greater than π΄πΉ plus π΅πΉ.