 # Investigation: Proof of the triangle inequality

Lesson

We will now present a complete paragraph proof of the triangle inequality theorem.

Reading proofs takes much longer than reading text written in everyday language, and that's okay! Follow these tips for understanding and reading proofs:

• Always stop reading if you don't understand something.  It's okay to go back and re-read a statement several times.
• If a writer draws something in a proof, try drawing it on your own with pencil and paper.
• After reading a few sentences, see if you can state back the meaning in your own words.

Consider a general triangle, $\Delta PQR$ΔPQR: What we want to prove now is that $PQ+QR>PR$PQ+QR>PR, $PQ+PR>QR$PQ+PR>QR, and $QR+PR>PQ$QR+PR>PQ.

It seems daunting to prove three things! But we begin by assuming that $\overline{PQ}$PQ is the longest side (as it is in the diagram above). Even if two or more of the sides have the same length as $\overline{PQ}$PQ, we are only assuming $PQ\ge QR$PQQR and $PQ\ge PR$PQPR.

This gives us two of the inequalities essentially for free. $PQ\ge PR$PQPR becomes $PQ+QR\ge PR+QR$PQ+QRPR+QR by the addition property of (in)equality, and since $QR$QR is a nonzero amount, we can be sure that $PQ+QR>PR$PQ+QR>PR on its own. A similar calculation establishes the second inequality.

So now we are left with the third inequality, which is going to take the most work. We are going to describe a construction and use it on a particular diagram, but we need to make sure it will work for any triangle and be careful as we proceed.

We perform the following steps:

1. Draw a circle of radius $QR$QR centered at $R$R.
2. Extend $\overline{PR}$PR to $\overrightarrow{PR}$PR.
3. The ray from step 2 intersects the circle from step 1 at a point - we mark it and call $X$X.

Here is the results of the process with our triangle: As indicated on the diagram, the radii $\overline{RX}$RX and $\overline{QR}$QR are congruent, since they're radii of the same circle, so they have the same measure. By substitution, $QR+PR>PQ$QR+PR>PQ is equivalent to $PR+RX>PQ$PR+RX>PQ, and proving this second statement is the same as finishing the proof altogether.

Remove the ray and the circle and create $\overline{XQ}$XQ to form the triangle $\Delta RQX$ΔRQX: As we noted before, two of the sides of this triangle are congruent, and so this triangle is isosceles. Since base angles of an isosceles triangle are equal, we can write $\angle RXQ\cong\angle RQX$RXQRQX.

Now we consider the other triangle we formed, $\Delta PQX$ΔPQX: The angle addition postulate tells us that $m\angle PQR+m\angle RQX=m\angle PQX$mPQR+mRQX=mPQX. If we subtract $m\angle PQR$mPQR from the left-hand side only, we break equality, and have the inequality $m\angle RQXmRQX<mPQX. Using the established congruence$\angle RXQ\cong\angle RQX$RXQRQX, we can substitute the measure of$\angle RXQ$RXQ for the measure of$\angle RQX$RQX and produce$m\angle RXQmRXQ<mPQX. Now $\angle PXQ=\angle RXQ$PXQ=RXQ since $P$P, $R$R, and $X$X all lie on the ray $\overrightarrow{XR}$XR, and so by substitution again we can write $m\angle PXQmPXQ<mPQX. We can then conclude that$PQPQ<PX, as an angle of smaller measure is opposite the shorter side.

By the segment addition postulate we know that $PX=PR+RX$PX=PR+RX, and by substituting this into the last inequality we found in the previous paragraph we know that $PQPQ<PR+RX. We once again use the fact that$RX=QR$RX=QR and by substitution we have$PQPQ<PR+QR. We can then rearrange this inequality to produce $QR+PR>PQ$QR+PR>PQ, QED.

#### Extra Challenge

Now that you've read the paragraph proof of the triangle inequality theorem, how can you be sure you really understand it?

Try re-creating the proof in a new medium.  For example, by explaining it in a video, cartoon strip, or as a two-column proof.

### Outcomes

#### GEO-G.CO.10

Prove and apply theorems about triangles.