Pennsylvania Geometry - 2020 Edition
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7.03 Inequalities in one triangle
Lesson

The measure of angles and length of sides in a triangle are related in a very useful way. It's true of every triangle you have ever seen, and when we state it, it might seem obvious. But just like other areas of geometry, we will carefully state the observation, and prove it is always true.

Exploration

The triangle $\Delta PQR$ΔPQR has side lengths $PQ=3$PQ=3, $QR=5$QR=5, and $PR=6$PR=6. Which of $\angle PQR$PQR, $\angle PRQ$PRQ, and $\angle QPR$QPR has the largest measure?

The best thing for us to do first is to draw this triangle. We want to get it exactly right, so we place vertex P on the page, and draw circles of radius 3 and 6 centered at P:

We choose any point we like on this first circle to be $Q$Q, to ensure the length of $\overline{PQ}$PQ is $3$3- we draw that segment in too. Since $\overline{QR}$QR has a length of $5$5, we also draw a circle of radius $5$5 centered at $Q$Q to get this diagram:

We then mark its intersection point with the circle of radius $6$6 centered at $P$P we drew earlier, and form a triangle from the resulting points:

Each of the side lengths match up correctly, and we have produced a triangle perfectly to scale. Now to answer our original question we remove the construction circles and just look at the triangle:

The angle with the largest measure is $\angle PQR$PQR. This angle looks close to being a right angle, while the other two are obviously smaller. Notice that this angle is opposite the longest side.

Now let's try a similar problem, but work in the other direction.

Consider $\Delta ABC$ΔABC , where $m\angle ABC=58^\circ$mABC=58°, $m\angle ACB=80^\circ$mACB=80°, and $m\angle BAC=42^\circ$mBAC=42°. Which of $\overline{AB}$AB, $\overline{AC}$AC, and $\overline{BC}$BC is the longest?

We draw the triangle by first producing two rays, $\overrightarrow{AX}$AX and $\overrightarrow{AY}$AY, that form an angle of $42^\circ$42° between them:

Now we need to select a point on each ray to form the correct angles at the other two vertices. Since we don't know the absolute measure of any side, we can select any point on the ray $\overrightarrow{AX}$AX to be $B$B, and consider the different segments formed by connecting $B$B to a point on $\overrightarrow{AY}$AY:

Each segment from $B$B to a point on $\overrightarrow{AY}$AY forms an angle ranging in measure from $0^\circ$0° to $180^\circ$180°, and we just need to pick the right one:

Which is the longest side? It looks like $\overline{AB}$AB. To check that each other side is shorter, we draw circles at each end passing through $C$C .

So the converse of our observation holds true here - the segment with the largest measure is $\overline{AB}$AB, and it is opposite the largest angle.


Now that we have some idea of what we're trying to show, let's see how it works in general.

Side-angle ordering theorem

For any triangle, the ordering of sides by length coincides with the ordering of the opposite angles by measure.  Therefore, the shortest side is opposite the smallest angle, and the longest side is opposite the largest angle.

In other words,

  • If $\overline{AX}$AX is longer than $\overline{AM}$AM, then $\angle AMX$AMX has larger measure than $\angle AXM$AXM, and
  • If $\angle AMX$AMX has a larger measure than $\angle AXM$AXM, then $\overline{AX}$AX is longer than $\overline{AM}$AM.

The key insight is contained in this diagram, where the triangle $\Delta AXM$ΔAXM has been reflected across the line bisecting $\angle MAX$MAX:

The two new points created, $N$N and $Y$Y, are then each part of an isosceles triangle - one is $\Delta MAN$ΔMAN, and the other is $\Delta XAY$ΔXAY. These two isosceles triangles are actually similar (their have two angles that are congruent), and we can use the properties of isosceles triangles as the basis of a formal proof of the theorem. Try using this applet to familiarize yourself with the idea:

This allows us to order the sides and angles in a triangle from smallest measure to largest measure without ever having to draw the triangle. There are also other results that rely on this statement always being true, and we will see them soon.

Outcomes

CC.2.3.HS.A.3

Verify and apply geometric theorems as they relate to geometric figures.

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