# What are strategies for teaching that the altitude of a right triangle creates two similar triangles?

If you draw the altitude to the right triangle as shown, it is easily seen that $$\triangle KLM\sim\triangle KNL\sim\triangle LNM.$$ This in turn leads to several interesting proportional relations like $$\frac{KN}{KL}=\frac{KL}{KM} \qquad\text{and}\qquad \frac{KN}{LN}=\frac{LN}{NM}.$$ These turn out to be crucial to what is arguably the canonical geometric proof of the Pythagorean theorem.

(Quick version: since $$AC^2=AD\cdot AB$$ and $$BC^2=BD\cdot AB$$, the two red regions have the same area and the two green regions have the same area.)

I'm looking for strategies to help my students master this concept. Is there a special name for this key geometric observation that would help my search, like how Thales' theorem was important enough to get its own name? My personal experience, even as a gifted student, is that it can be hard to see the similarity when you have to reflect one of the triangles to visualize the dilation compared to a similarity diagram with a triangle and one of its midsegments, for instance.

• If it is a proposition in Euclid's Elements, then you could refer to it by its number. Feb 25 '20 at 13:40
• @GeraldEdgar True, but that wouldn't help me with searching YouTube or teacherspayteachers for resources. Anyway, <blush> I just looked it up in Elements and found that that wasn't actually how Euclid proved the Pythagorean theorem after all, even though it was the proof given in my Euclid-centric geometry textbook back in the 80's. I edited my title to make it clear that I'm looking for effective teaching strategies more than a name I can Google for. Feb 25 '20 at 14:17
• It has been commented the Euclid's proof of the Pythagorean theorem is done as a "tour de force" in Book I, instead of waiting until later after the discussion of similarity, which would make for an easier proof. Feb 25 '20 at 17:47
• I don't know about name, but for a quick proof: two right angled triangles are clearly similar if they share a non-right angle. Of the three triangles (big, left, right): big and left share the angle at K, and big and right share the angle at M, so they're all similar. Does that help with the explanation at all? Feb 26 '20 at 12:29
• It is not entirely clear what you are asking here: the text of your question asks for a name for this statement, while the title asks more generically for "teaching strategies". Feb 26 '20 at 12:38

I've always seen it referred to as the "Right Triangle Altitude Theorem".

It may take a bit of practice, but I start by declaring $$\angle M$$ and $$\angle K$$ complementary. Then I draw a mark to indicate $$\angle NLM$$ and show that it is also complementary to $$\angle M$$. Thus $$\angle NLM$$ is congruent to $$\angle K$$ and so on.

I do this visually by getting 2 large pieces of white cardboard, and draw the triangle as you show, but then have a second layer taped on top, same triangle, but cut along line LN. So I can remove the two smaller triangles and then orient all 3 to the visually pleasing 'right angle on bottom view'.

• That's great! I've wanted a manipulative for this, but I've always imagined higher tech than two pieces of cardboard and a piece of poster board. But, really, that's all I'd need to drive the lesson home! Feb 25 '20 at 14:19
• I started to type up an answer, but the more I typed, the more I realized it was similar (heh) to JTP's answer. I did find that there's a name for this technique, the Similarity Test AAA. Really, it's a matter of (a) reminding them that the sum of the angles of a triangle is always the same (180° if you must), reminding them that the altitude creates two more right triangles, and (c) demonstrating that each of the smaller triangles shares an angle with the larger triangle, so (considering the right triangle) the third angle must be equal as well. Feb 26 '20 at 4:02

You could present the concept that, ignoring scale, a right triangle can be defined entirely by a single number: its smallest interior angle. Call this the triangle's "describing number." If you want to dramatize this, you could specify some angle like 30 degrees and ask students to cut triangles out of paper that have this describing number. Comparing all the triangles, it will be obvious that they're all similar, but of different sizes. Some may have to be flipped over.

In the proof you're presenting, where you dissect a right triangle into two smaller right triangles, you could ask where those triangles get their describing numbers. Clearly they can only depend on the parent triangle's describing number. This is not by itself a proof that their describing numbers are equal to that of the parent, but when you then prove that they are, I don't think it should be surprising at all.

• I like this. You could perhaps generalize that you can consider either of the interior angles the "defining number" because it specifies the entire "similar triangles set" it belongs to, in particular it defines the other angle. It then follows immediately that the smaller (top) angle of the left triangle NLK is defined by the angle at K, and hence must be the same as the angle at M, which is defined by the same K angle -> the triangles are similar. (I suppose it's the same principle at work in all other answers but calling the angle "describing number" made me understand it.) Feb 27 '20 at 10:31

If you have established that the sum of the interior angles of a triangle is always 180°, you're basically done. Each of the small triangles shares one angle with the original one; the second angle is a right angle. So the third one needs to be the same as the "missing" angle. All three angles being the same, the triangles have to be similar.

The similitude is stated in Euclid's book VI, proposition 8.

That is not exactly the same statement, but two metric relations that are equivalent to those similitudes are known in Italy as Euclid's first and second theorem:

So there would be at least some historical motivation to call that similitude Euclid's theorem, even though the name is already taken for the statement (in a completely different field) that there are an infinite number of primes.

At secondary school level, label one angle 'x' and work your way around the shape labelling each angle relative to the first one you chose. You'll get a variety of 'x' and '90-x' angles. Then a concluding sentence to say that the angles correspond and so the triangles are similar.