 Section 5.5 is all about inequalities that involve two triangles. Pull out your notes and get this information into your table of contents. So the first theorem we'll discuss is called the Hinge Theorem, also known as the SAS inequality. Hinge Theorem states, in two triangles, if you have two sets of sides that are congruent and the included angles are not congruent, then the third sides are also not equal in the same order that those two angles are not congruent. In other words, if we have these two triangles and we're given that AB is congruent to DE, AC is congruent to DF, and then we learn that angle A is bigger than the measure of angle D, we can prove that BC is longer than EF. So that's the hinge theorem, and the converse is also true. The converse states, if two sets of sides are congruent and the third pair of sides are not congruent, then the angles opposite those sides are not congruent as well. In other words, if we have AB congruent to DE, AC congruent to DF, but BC is longer than EF, the converse of the hinge theorem allows us to prove that angle A is larger in measure than the measure of angle D. So a couple of examples for you. Here we are asked to compare the lengths of WX and VY. So WX we see is kind of the shorter side of the triangle on top. VY is the side of the bottom triangle that's unlabeled. We see 15 is congruent to 15. We see we've got a shared side, VX, and then angle 20 of course is less than angle 50. And so therefore the side opposite 20 is going to be shorter than the side opposite 50. This is using the side angle side inequality theorem because we know two pairs of sides, the 15s and the shared side, and we know information about the non-congruent angles. So therefore the sides opposite the angles must be non-congruent. So side angle side implies that WX is less than VY. Here's another example. Compare the angles ABD and CBD. Well just like before we've got a pair of congruent sides. In this particular drawing we've got a shared side, BD, and then we notice that 4 is clearly less than 8, and so therefore the angle opposite 4 must be smaller than the angle opposite 8. And so therefore by side side side inequality we're using side side side because we know information about all the sides and we're making a conclusion about the angles. We can therefore conclude that the measure of angle ABD is less than the measure of angle CBD. Next take a look at this example. Our job is to write and solve an inequality to find X. Here we have three triangles although we don't really need the middle isosceles triangle. That middle isosceles is only giving us information about those green congruent sides. So we have those greens, we have the 250s, and we have the blue angles. So therefore using side angle side inequality we can make a statement like 2x minus 8 must be smaller than x plus 2 because angle 84 degrees is less than 110 degrees. So let's set up an inequality. Since 84 degrees is less than 110 degrees, I know 2x minus 8 must be less than x plus 2. And so if we go through and solve this we see that x must be less than 10. Now the statement x is less than 10. We could be talking about 9 or 8 or 6, but we could also be talking about negative 500. So there are some implied restrictions when dealing with these triangle inequalities. What I mean by that is we can't have segment lengths less than 0. So therefore the smaller of the two segments that we don't know, 2x minus 8, it can't be less than 0. In fact it has to be greater than 0 in order for the triangle to exist. So therefore 2x minus 8 has to be greater than 0, and so x must be greater than 4. And so we combine these two inequalities, x greater than 4, x less than 10, and we finalize with x is greater than 4 and less than 10. Another example, here we have another right and solve an inequality to find x. We're given that 2 side lengths are 52, we have a shared side, and we have those green sides 42 and 12. So we have information on three pairs of sides, and we want to solve for angles. So we know that 33 must be the larger of the two angles. So 33 must be bigger than 3x minus 3 because of the side-side-side inequality theorem. So since 32 is bigger than 12, I know 3x minus 3 must be smaller than 33. So therefore x must be less than 12. And once again, just like the previous example with segments, angles can't be negative. However, we've just shown that x must be less than 12. So x could be negative 500. So we need some implied restrictions. Interior angles of triangles must be greater than 0, but also less than 180 degrees. So if we solve these two inequalities, we see that x must be greater than 1 and x must be less than 61. Wait a minute. x less than 61. We were already given or we already found that x must be less than 12. So that's kind of a ridiculous restriction. So to put it another way, we know that x must be less than 12 and greater than 1, but less than 61. The only overlap here, where all three of these restrictions overlap, is between 1 and 12. The fact that x is less than 61 is true, but it's not really helpful for us in this case. So our final answer is that x is less than 12 and greater than 1. Now we have a few proof examples. In this figure, we're given that gl is less than dl, that ga is equal in length to ad, and our task is to prove that the measure of angle 1 is less than the measure of angle 2. So let's start by labeling our given information. We know that, excuse me, ga is equal to ad, so I've got the green sides there. I've got my shared side al, marked in red, and then the unequal sides I have highlighted in orange, the smaller side and the bigger side. So with information about the three sides, and our job is to prove something about angles, we're going to use the side-side-side inequality. Side-side-side because we know information about all three sides. So side-side-side inequality matches chapter 4 proofs. We have one pair of sides congruent, we have another pair of sides congruent, and we had one given pair of sides unequal in measure. So since gl was smaller than dl, I know angle 1 must be less than angle 2. And there we go. Pretty straightforward proof. We have one more proof example. In this example, bd is a median of triangle abc, and our task is to prove that the mid... pardon, we're also given that a is greater than c, and our task is to prove that 1 is less than... pardon, is greater than 2. So to prove that angle 1 is greater than 2, first off, we see medians, of course, gives us that d is a midpoint, and so ad is congruent to cd. We have our shared side, db. But in order to prove that angle 1 is congruent, pardon, is greater than 2, we need to somehow prove that angle, or that side length ab is less than bc. How can we do that? Well, that second piece of given information, that angle a is greater than c, that's gonna help us out. Let's take a look at, oops, I mislabeled, that should be c, that should be b. Let's take a look at triangle abc, the entire triangle. We see a is greater than c, and so therefore the side opposite a is greater than the side opposite c. That harkens back to some of our previous triangle theorems, where in one triangle, if angles are unequal, then sides opposite those angles are unequal. And so that allows us to prove that ab is less than bc. So we're still using side, side, side. We know that d is the midpoint of ac by the midpoint definition, but d came from the fact that bd was a median of triangle abc. We knew that bd was shared, and finally that a was bigger than c, which implies in one triangle, if the angles are not congruent, then the sides opposite those angles are not congruent. So that allowed us to prove that ab was less than bc. And there we go, angle 1 is greater than angle 2.