Geometry may seem like no laughing matter, but this lesson has more than one HA moment. That's because this is all about the Hypotenuse Angle Theorem, or HA Theorem, which allows you to prove congruence of two right triangles using only their hypotenuses and acute angles.

Before we start roaring with all the laughs this lesson brings (!), let's make sure we have a firm understanding of **right triangles**. A right triangle is called that because it has one right angle ($90\xb0$), which means the other two angles must be acute (less than $90\xb0$).

This right angle limits the possible measurements of the other two angles. Together they must add up to $90\xb0$ because all interior angles of *any* triangle -- right, scalene, obtuse -- must add to $180\xb0$. Subtracting the one right angle from $180\xb0$ leaves only $90\xb0$ to be shared by the two remaining angles, making both of them acute angles.

Can you see what would happen if we knew something about one of those two acute angles? We would know something about the remaining angle. Let's take a look at how that plays a role in the HA Theorem.

Now that we have right triangles right in our heads, let's look at the HA Theorem.

The **HA Theorem** states; If the hypotenuse and an acute angle of a right triangle are congruent to the hypotenuse and an acute angle of another triangle, then the two triangles are congruent.

Congruence does not mean just somewhat alike; it means the two triangles will be identical; every side and every angle, equal between the triangles. That's a tall order, and we are claiming to get it just by knowing one side (the hypotenuse) and one angle.

Remember, though, that we already know a second angle. We know the right angle that forms a square corner.

Here are two right triangles, $\u25b3ZAP$ and $\u25b3HOT$.

They are both facing with their hypotenuses to the right, which means their right angles are to the left -- HA! *(A small touch of triangle humor.)*

Notice $\angle A$ and $\angle O$ are right angles, indicated by the little square $\square $ tucked into the interior angles. We are told that the hypotenuses, $ZP$ and $HT$, are congruent, which is why they have the little matching hash marks. We are also told that acute $\angle Z$ and $\angle H$ are congruent, shown by their own hash marks.

If we knew only that much geometry, we would be stumped. We could say the six parts (three sides and three angles) have only three parts congruent, and they are not all touching.

Look carefully -- $\angle A$ and $\angle Z$ are consecutive angles in our left right triangle … uh … our right triangle on the left *(HA humor again)*. Those two angles do not include a known side between them. We have no idea if $ZA$ is congruent to $HO$.

Check out the remaining angles. $\angle P$ and $\angle T$. What do we know about them? We know they are congruent. Why?

They must be congruent because of what we said earlier. **Given two of the angles, the third angle is found by subtracting the two given angles from $180\xb0$**. We do not even need numbers for $\angle Z$ and $\angle H$; they are congruent, so $\angle P$ and $\angle T$ are congruent.

So what, you say? If we know that all three angles are congruent, and we know that included sides between angles are congruent, then we have the ASA Postulate! Recall that ASA tells us:

Triangles are congruent if any two angles and their included side are equal in the triangles.

Building off that handy right angle, we worked out two included angles, on either side of the hypotenuse. Now we have all these congruences:

- $\angle A\cong \angle O$ (two right angles, which we used to deduce $\angle P\cong \angle T$)
- $\angle Z\cong \angle H$ (a given)
- Hypotenuse $ZP$ ≅ hypotenuse HT (a given)
- $\angle P\cong \angle T$ (deduced from $\angle Z\cong \angle H$ and $\angle A\cong \angle O$)

The last three congruences are the ASA Postulate at work. HA! We did some amazing detective work there.

Do you need to go through all that every time you want to show two right triangles are congruent? No. You can use the HA Theorem! HA! *(We told you this would have more than one HA moment.)*

Instead of going through the lengthy process of finding the third angle congruent, hauling out the ASA Postulate, and declaring the two right triangles congruent, you can easily apply the HA Theorem.

You cannot show off the HA Theorem with something as simple as two twin right triangles, charming as $\u25b3ZAP$ and $\u25b3HOT$ are. What about something trickier, like two right triangles seeming to slide past each other, like these:

These two right triangles were constructed from line $OA$, intersected by line $FB$, crossing at $PointG$.

Right $\u25b3FOG$ shares a vertex, $PointG$, with $\u25b3BAG$. We see that $\angle O$ and $\angle A$ are right angles, and the little hash marks tell us hypotenuses $FG$ and $BG$ are congruent. What else are we told? Nothing!

Are you ready to have a HA moment? We know sides $OG$ and $AG$ form a straight line, because they are segments of line $OA$. We know that both right triangles share $PointG$, creating two interior angles ($\angle FGO$ and $\angle BGA$). Those interior angles are vertical angles of two crossing lines! HA! Vertical angles are congruent. Now we have another set of congruences. Let's make a list:

- $\angle FGO\cong \angle BGA$
- $HypotenuseFG\cong HypotenuseGB$

With just the hypotenuse and one acute angle, we now release the power of the HA Theorem and state that:

$\angle FOG\cong \angle BAG$

Though it may not have been a barrel of laughs, by exploring the HA Theorem you are now able to recall and state the **Hypotenuse Angle (HA) Theorem**, demonstrate the HA Theorem's connection to the Angle Side Angle Theorem, and mathematically prove the HA Theorem.

After studying these instructions, illustrations and the video, you will be able to:

- Recall and state the Hypotenuse Angle (HA) Theorem
- Demonstrate the HA Theorem's connection to the Angle Side Angle Theorem
- Mathematically prove the HA Theorem

Instructor: **Malcolm M.**

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

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