Isosceles triangles have equal legs (that's what the word "isosceles" means). Yippee for them, but what do we know about their base angles? How do we *know* those are equal, too? We reach into our geometer's toolbox and take out the Isosceles Triangle Theorem. No need to plug it in or recharge its batteries -- it's right there, in your head!

Here we have on display the majestic isosceles triangle, $\u25b3DUK$. You can draw one yourself, using $\u25b3DUK$ as a model.

Hash marks show sides $\angle DU\cong \angle DK$, which is your tip-off that you have an isosceles triangle. *If* these two sides, called **legs**, are equal, *then* this is an isosceles triangle. What else have you got?

Let's use $\u25b3DUK$ to explore the parts:

- Like any triangle, $\u25b3DUK$ has three interior angles: $\angle D$, $\angle U$, and $\angle K$
- All three interior angles are acute
- Like any triangle, $\u25b3DUK$ has three sides: $DU$, $UK$, and $DK$
- $\angle DU\cong \angle DK$, so we refer to those twins as legs
- The third side is called the
**base**(even when the triangle is not sitting on that side) - The two angles formed between base and legs, $\angle DUK$ and $\angle DKU$, or $\angle D$ and $\angle K$ for short, are called
**base angles**:

Knowing the triangle's parts, here is the challenge: how do we *prove* that the base angles are congruent? That is the heart of the **Isosceles Triangle Theorem**, which is built as a conditional *(if, then)* statement:

The **Isosceles Triangle Theorem** states: *If* two sides of a triangle are congruent, *then* angles opposite those sides are congruent.

To mathematically prove this, we need to introduce a median line, a line constructed from an interior angle to the midpoint of the opposite side. We find $PointC$ on base $UK$ and construct line segment $DC$:

There! That's just $DUCKy$! Look at the two triangles formed by the median. We are given:

- $UC\cong CK$ (median)
- $DC\cong DC$ (reflexive property)
- $DU\cong DK$ (given)

We just showed that the three sides of $\u25b3DUC$ are congruent to $\u25b3DCK$, which means you have the **Side Side Side Postulate**, which gives congruence. So if the two triangles are congruent, then corresponding parts of congruent triangles are congruent (CPCTC), which means …

- $\angle U\cong \angle K$

The converse of a conditional statement is made by swapping the hypothesis *(if …)* with the conclusion *(then …)*. You may need to tinker with it to ensure it makes sense. So here once again is the Isosceles Triangle Theorem:

Iftwo sides of a triangle are congruent,thenangles opposite those sides are congruent.

To make its converse, we *could* exactly swap the parts, getting a bit of a mish-mash:

Ifangles opposite those sides are congruent,thentwo sides of a triangle are congruent.

That is awkward, so tidy up the wording:

The ** Converse of the Isosceles Triangle Theorem** states: *If* two angles of a triangle are congruent, *then* sides opposite those angles are congruent.

Now it makes sense, but is it true? Not every converse statement of a conditional statement is true. *If* the original conditional statement is false, *then* the converse will also be false. *If* the premise is true, *then* the converse could be true or false:

IfI see a bear,thenI will lie down and remain still.

IfI lie down and remain still,thenI will see a bear.

For that converse statement to be true, sleeping in your bed would become a bizarre experience.

Or this one:

IfI have honey,thenI will attract bears.

IfI attract bears,thenI will have honey.

Unless the bears bring honeypots to share with you, the converse is unlikely ever to happen. And bears are famously selfish.

To prove the converse, let's construct another isosceles triangle, $\u25b3BER$.

Given that $\angle BER\cong \angle BRE$, we must prove that $BE\cong BR$.

Add the angle bisector from $\angle EBR$ down to base $ER$. Where the angle bisector intersects base $ER$, label it $PointA$.

Now we have two small, right triangles where once we had one big, isosceles triangle: $\u25b3BEA$ and $\u25b3BAR$. Since line segment $BA$ is an angle bisector, this makes $\angle EBA\cong \angle RBA$. Since line segment $BA$ is used in both smaller right triangles, it is congruent to itself. What do we have?

- $\angle BER\cong \angle BRE$ (given)
- $\angle EBA\cong \angle RBA$ (angle bisector)
- $BA\cong BA$ (reflexive property)

Let's see … that's an angle, another angle, and a side. That would be the **Angle Angle Side Theorem**, AAS:

The **AAS Theorem** states: *If* two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, *then* the triangles are congruent.

With the triangles themselves proved congruent, their corresponding parts are congruent (CPCTC), which makes $BE\cong BR$. The converse of the Isosceles Triangle Theorem is true!

By working through these exercises, you now are able to recognize and draw an isosceles triangle, mathematically prove congruent isosceles triangles using the **Isosceles Triangles Theorem**, and mathematically prove the converse of the Isosceles Triangles Theorem. You also should now see the connection between the Isosceles Triangle Theorem to the Side Side Side Postulate and the Angle Angle Side Theorem.

After working your way through this lesson, you will be able to:

- Recognize and draw an isosceles triangle
- Mathematically prove congruent isosceles triangles using the Isosceles Triangles Theorem
- Mathematically prove the converse of the Isosceles Triangles Theorem
- Connect the Isosceles Triangle Theorem to the Side Side Side Postulate and the Angle Angle Side 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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