Triangles and Congruency

Classifying Triangles

Later we’ll talk about similar triangles here, altitudes, medians, and angle bisectors of triangles here, and areas of triangles here.

A triangle is a three-sided polygon and has three angles.

A triangle can be classified by angles:

- An acute triangle has three acute angles $( \displaystyle <90{}^\circ )$.
- An obtuse triangle has one obtuse angle $( \displaystyle >90{}^\circ )$.
- A right triangle has one right angle $(90{}^\circ )$.

Here are examples:

Triangles are also classified by sides:

- - An equilateral triangle has three congruent sides, as well as three congruent angles.
  - An isosceles triangle has at least two sides that are congruent, with congruent angles oppositive the sides. Note that equilateral triangles are also isosceles.
  - A scalene triangle has no two sides that are congruent.

Here are examples:

Angles of Triangles

You may have learned by now that the sum of the three angles in a triangle is equal to $ \displaystyle \boldsymbol{180{}^\circ} $; this is because the angles of a triangle can be rearranged to form a straight line. The formal reason for this is the Triangle Sum Theorem, which also states that if the measures of two angles of a triangle are known, the measure of the third angle can be determined. This is simple math, right?

From this, we can prove the Exterior Angle Theorem, which states that the measure of an exterior angle of at triangle (angle on the outside formed by a side and an extension of an adjacent side) is equal to the sum of the measures of the two interior angles that are not adjacent (called remote interior angles):

$ \displaystyle m\angle 1 +m\angle 2=m\angle 3 $

Two corollaries come from this theorem (using simple addition or subtraction):

- - - Acute angles of a right triangle are complementary (their sum equals $\displaystyle 90{}^\circ )$.
    - A triangle can have at most one right or obtuse angle.

There are a few Triangle Inequality Theorems:

- - - For unequal sides of a triangle, the greater angles are opposite the greater sides, and the smaller angles are opposite the smaller sides. Try this yourself using scissors and your hand to created angles of a triangle.
    - The sums of lengths of any two sides in a triangle is greater than the length of the third side. Try this by drawing different triangles and noting the lengths of sides.

Congruent Triangles

Triangles with the same shape and size are congruent. In other words, two triangles are congruent if the three sides and angles are the same as the corresponding three sides and angles of the other triangle. Corresponding parts refer to the parts (the three sides and three angles) that are congruent between the congruent triangles; we can mark them to show which of the parts have the same measurement and thus are corresponding/congruent:

Note that it’s very important to label and name the triangles correctly; the angles of the first triangle must line up with the respective congruent angles of the second triangle. Thus, in the example above, $ \displaystyle \vartriangle ABC\cong \vartriangle DEF$, but $ \displaystyle \vartriangle ABC$ is not necessarly $ \cong$ to $ \vartriangle DFE$.

There are certain tests for congruency for triangles, so we don’t have to show that all six corresponding parts are congruent. Here are some of the postulates we can use:

Postulate Name	Explanation	Example
SSS Postulate (Side-Side-Side Postulate)	If three sides of one triangle are congruent to three sides of another, the two triangles are congruent.	$ \displaystyle \overline{{AB}}\cong \,\overline{{DE}},\,\,\overline{{AC}}\cong \,\overline{{DF}},\,\,\overline{{BC}}\cong \,\overline{{EF}}$ $ \displaystyle \vartriangle ABC\cong \vartriangle DEF$ Do you see how the sides line up with the order of the triangle vertices? For example, for $ \displaystyle \overline{{AB}}\cong \,\overline{{DE}}$, these are the respective first two letters of the triangle congruence statement.
SAS Postulate (Side-Angle-Side Postulate)	If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, the triangles are congruent. Note: the congruent angles have to be inside the two congruent sides; think of going around the triangle without skipping an angle or side.	$ \displaystyle \overline{{AB}}\cong \,\overline{{DE}},\,\,\angle B\cong \angle E,\,\,\,\overline{{BC}}\cong \overline{{EF}}$ $ \displaystyle \vartriangle ABC\cong \vartriangle DEF$
AAS Postulate (Angle-Angle-Side Postulate)	If two angles and the opposite (non-included) side of a triangle are congruent to two angles and the opposite side of another triangle, the triangles are congruent.	$ \displaystyle \angle A\cong \angle D,\,\,\,\angle C\cong \angle F,\,\,\,\overline{{AB}}\cong \overline{{DE}}$ $ \displaystyle \vartriangle ABC\cong \vartriangle DEF$
HL Postulate (Hypotenuse-Leg Postulate)	If the hypotenuse and another leg of a right triangle are congruent to these corresponding parts of another triangle, the triangles are congruent. Note that the hypotenuse of a right triangle is the side that is opposite the right angle.	$ \displaystyle \angle B,\,\,\angle E$ are right angles. $ \displaystyle \overline{{AB}}\cong \,\overline{{DE}}$ (leg), $\overline{{AC}}\cong \,\overline{{DF}}$ (hypotenuse) $ \displaystyle \vartriangle ABC\cong \vartriangle DEF$

Postulate Name

Explanation

Example

SSS Postulate

(Side-Side-Side Postulate)

If three sides of one triangle are congruent to three sides of another, the two triangles are congruent.

$ \displaystyle \overline{{AB}}\cong \,\overline{{DE}},\,\,\overline{{AC}}\cong \,\overline{{DF}},\,\,\overline{{BC}}\cong \,\overline{{EF}}$

$ \displaystyle \vartriangle ABC\cong \vartriangle DEF$

Do you see how the sides line up with the order of the triangle vertices? For example, for $ \displaystyle \overline{{AB}}\cong \,\overline{{DE}}$, these are the respective first two letters of the triangle congruence statement.

SAS Postulate

(Side-Angle-Side Postulate)

If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, the triangles are congruent.

Note: the congruent angles have to be inside the two congruent sides; think of going around the triangle without skipping an angle or side.

$ \displaystyle \overline{{AB}}\cong \,\overline{{DE}},\,\,\angle B\cong \angle E,\,\,\,\overline{{BC}}\cong \overline{{EF}}$

$ \displaystyle \vartriangle ABC\cong \vartriangle DEF$

AAS Postulate

(Angle-Angle-Side Postulate)

If two angles and the opposite (non-included) side of a triangle are congruent to two angles and the opposite side of another triangle, the triangles are congruent.

$ \displaystyle \angle A\cong \angle D,\,\,\,\angle C\cong \angle F,\,\,\,\overline{{AB}}\cong \overline{{DE}}$

$ \displaystyle \vartriangle ABC\cong \vartriangle DEF$

HL Postulate

(Hypotenuse-Leg Postulate)

If the hypotenuse and another leg of a right triangle are congruent to these corresponding parts of another triangle, the triangles are congruent.

Note that the hypotenuse of a right triangle is the side that is opposite the right angle.

$ \displaystyle \angle B,\,\,\angle E$ are right angles.

$ \displaystyle \overline{{AB}}\cong \,\overline{{DE}}$ (leg), $\overline{{AC}}\cong \,\overline{{DF}}$ (hypotenuse)

$ \displaystyle \vartriangle ABC\cong \vartriangle DEF$

Note that, in these cases, the triangles are not necessarily congruent if they have corresponding congruent parts: AAA (all angles are congruent), SSA (two adjacent sides, followed by an adjacent angle are congruent).

Congruent Parts of Triangles

Again, the parts of triangles that correspond between triangles are called corresponding parts. It makes sense that, after proving triangle congruence, the corresponding parts of congruent triangles would be congruent. Thus we have the CPCTC Theorem: Corresponding Parts of Congruent Triangles are Congruent. You’ll see this a lot, since it doesn’t depend on how the triangles were proven congruent. Thus, if $ \displaystyle \vartriangle ABC\cong \vartriangle DEF$, we have:

$ \displaystyle \overline{{AB}}\cong \overline{{DE}},\,\,\,\overline{{BC}}\cong \overline{{EF}},\,\,\,\overline{{AC}}\cong \overline{{DF}}$ and $ \displaystyle \angle A\cong \angle D,\,\,\angle B\cong \angle E,\,\,\angle C\cong \angle F$:

Just think of CPCTC as “picking apart” congruent triangles’ parts after proving the congruence.

Isosceles Triangles

Isosceles Triangles are triangle with two congruent sides, which are called the legs. The angle at the “top” of the legs is called a vertex angle, and the angles attaching the legs to the base are called the base angles. Thus, in the diagram, angle A is the vertex angle, and angles B and C are the base angles.

Here are some properties of isosceles triangles, that can be proven:

- - If two sides of a triangle are congruent, then the angles opposite those sides are congruent. (Isosceles Triangle Theorem)
  - If two angles of a triangle are congruent, then the sides opposite those angles are congruent. (Converse of Isosceles Triangle Theorem)
  - If two sides of a triangle are congruent to two sides of a second triangle and the included angle of the first triangle is larger than the included angle of the second triangle, the third side of the first triangle is longer than the third side of the second triangle. The converse is also true. Makes sense, right? (Hinge Theorem)

Problems and Solutions

Problem: Find $ x$ and the measure of $ \displaystyle \angle \,1$:

Solution: By the Exterior Angle Theorem, we know that $ (8x+120)=(12x+20)+(16x+40)$. Solving, we get $ x=3$. $ \displaystyle \angle \,1$ is supplementary to the exterior angle by the supplementary theorem (they form a linear pair), which is represented by $ 8x+120$; thus, $ \displaystyle 180-\left[ {8(3)+120} \right]=180-144=36{}^\circ $. Try the other angles; the interior angles add up to $180{}^\circ $!

Problem: List the sides in order from smallest to largest (drawing not drawn to scale):

Solution: Sides opposite smaller angles are smaller; sides opposite larger angles are larger; the converses are also true. The sides in order from smallest to largest are $ \displaystyle \overline{{BC}}$, $ \displaystyle \overline{{AC}}$, $ \displaystyle \overline{{AB}}$.

Problem: Find the range of values the third side of a triangle can have, given these two sides: $ 12$ feet and $ 18$ feet.

Solution: Let $ s$ be the measurement of the unknown side. We know each of two sides of a triangle must be greater than the third (a Triangle Inequality Theorem), so we first have $ 12+18>s$. But do you see how this third side $ s$ must also be greater than the difference of the other two sides: $ 18-12<s$? If this weren’t the case, $ s+12$ wouldn’t be greater than $ 18$. Thus, $ 18-12<s<12+18$, or $ 6<s<30$. Try it; it works!

Problem: Using a flow proof, prove the Exterior Angle Theorem (measure of an exterior angle of at triangle is equal to the sum of the measures of the two remote interior angles):

Given: Triangle ABC:

Prove: $ \displaystyle m\angle 1 +m\angle 2=m\angle 4 $

Note: A flow proof is a proof using diagrams, either vertical or horizontal, to show the process of the proof.

Solution: Here is a vertical flow chart that would work!

Problem:

Given: $ \displaystyle \overline{{AB}}\cong \overline{{DC}}$ and $ \displaystyle \overline{{AB}}\,\,||\,\,\overline{{DC}}$

Prove: $ \displaystyle \vartriangle DAB\cong \vartriangle BCD$

Solution:

Proof:

Statements	Reasons
1. $ \displaystyle \overline{{AB}}\cong \overline{{DC}}$	1. Given
2. $ \displaystyle \overline{{DB}}\cong \overline{{DB}}$	2. Reflexive
3. $ \displaystyle \overline{{AB}}\,\,\|\|\,\,\overline{{DC}}$	3. Given
4. $ \displaystyle \angle \,1\cong \angle \,2$	4. $ \displaystyle \text{AIA}\Rightarrow \,\cong \,\angle \,\text{ }\!\!’\!\!\text{ s}$
5. $ \displaystyle \vartriangle DAB\cong \vartriangle BCD$	5. $ \displaystyle \text{SAS}\cong \,\vartriangle \text{ }\!\!’\!\!\text{ s}$

See how new concepts build on the old concepts we’ve learned?!

Problem:

Given: $ \displaystyle \overline{{AC}}$ is a $ \displaystyle \bot $ bisector of $ \displaystyle \overline{{BD}}$

Prove: $ \displaystyle \overline{{AB}}\cong \overline{{AD}}$

Solution:

Proof:

Statements	Reasons
1. $ \displaystyle \overline{{AC}}$ is a $ \displaystyle \bot $ of $ \displaystyle \overline{{BD}}$	1. Given
2. $ \displaystyle \angle \,3$ and $ \displaystyle \angle \,4$ are right angles	2. $ \displaystyle \bot \,\,\Rightarrow \,\,\text{Right}\,\angle \text{ }\!\!’\!\!\text{ s}$
3. $ \displaystyle \angle \,3 \cong \angle \,4$	3. Right Angles are $ \cong $
4. $ \displaystyle \overline{{BC}}\cong \overline{{CD}}$	4. Def of Bisector
5. $ \displaystyle \overline{{AC}}\cong \overline{{AC}}$	5. Reflexive
6. $ \displaystyle \vartriangle ACB\cong \vartriangle ACD$	6. SAS
7. $ \displaystyle \overline{{AB}}\cong \overline{{AD}}$	7. CPCTC

Note: Even though we have two right triangles, we couldn’t use HL congruence since we didn’t have information about the hypotenuses.

Problem:

Given: $ \displaystyle \angle ABC$ and $ \displaystyle \angle BCD$ are right angles, $ \displaystyle \overline{{AC}}\cong \overline{{BD}}$

Prove: $ \displaystyle \angle \,1\cong \angle \,2$

Solution:

Proof:

Statements	Reasons
1. $ \displaystyle \angle ABC$ and $ \displaystyle \angle DCB$ are right angles	1. Given
2. $ \displaystyle \vartriangle ABC$ and $ \displaystyle \vartriangle DCB$ are right triangles	2. Def right triangle
3. $ \displaystyle \overline{{AC}}\cong \overline{{BD}}$	3. Given
4. $ \displaystyle \overline{{BC}}\cong \overline{{BC}}$	4. Reflexive
5. $ \displaystyle \vartriangle ABC\cong \vartriangle DCB$	5. Hypotenuse-Leg (HL)
6. $ \displaystyle \angle \,1\cong \angle \,2$	6. CPCTC

Here’s a long one; don’t let it discourage you! Just plug on through, knowing that there must be some congruent triangles involved! Note that there may be easier ways to do this proof, using some of the theorems we used:

Problem:

Given: $ \displaystyle \angle \,2\cong \angle \,3$, $ \displaystyle \angle \,7\cong \angle \,8$

Prove: $ \displaystyle \overline{{DF}}\cong \overline{{EF}}$

Solution:

Proof:

Statements	Reasons
1. $ \displaystyle \angle \,7\cong \angle \,8$	1. Given
2. $ \displaystyle \angle \,5$ and $ \displaystyle \angle \,7$ form linear pair, $ \displaystyle \angle \,6$ and $ \displaystyle \angle \,8$ form linear pair	2. Def of Linear Pair
3. $ \displaystyle \angle \,5$ and $ \displaystyle \angle \,7$ are supplementary, $ \displaystyle \angle \,6$ and $ \displaystyle \angle \,8$ are supplementary	3. Linear pair $ \displaystyle \Rightarrow \,\,\text{Supple}\,\angle \text{ }\!\!’\!\!\text{ s}$
4. $ \displaystyle m\angle \,5 + m\angle \,7=180$, $ \displaystyle m\angle \,6 + m\angle \,8=180$	4. Def of Supp
5. $ \displaystyle m\angle \,5 =180- m\angle \,7 $, $ \displaystyle m\angle \,6 =180- m\angle \,8 $	5. Subt Prop of =
6. $ \displaystyle m\angle \,7 = m\angle \,8$	6. Def of $ \cong$
7. $ \displaystyle m\angle \,5 =180- m\angle \,7 $, $ \displaystyle m\angle \,6 =180- m\angle \,7 $	7. Subst
8. $ \displaystyle m\angle \,5 = m\angle \,6$	8. Trans (or Subst)
9. $ \displaystyle \angle \,5\cong \angle \,6$	9. Def of $ \cong$
10. $ \displaystyle \overline{{AG}}\cong \overline{{AG}}$	10. Reflexive
11. $ \displaystyle \angle \,2\cong \angle \,3$	11. Given
12. $ \displaystyle \vartriangle AGD\cong \vartriangle AGE$	12. ASA
13. $ \displaystyle \overline{{DG}}\cong \overline{{GE}}$	13. CPCTC
14. $ \displaystyle \overline{{GF}}\cong \overline{{GF}}$	14. Reflexive
15. $ \displaystyle \vartriangle DGF\cong \vartriangle EGF$	15. SAS
16. $ \displaystyle \overline{{DF}}\cong \overline{{EF}}$	16. CPCTC

See how we used the Definition of Congruence to turn Geometry into “math” and then back to Geometry?

Problem:

Given: $ \displaystyle \overline{{AC}}$ is a $ \displaystyle \bot $ bisector of $ \overline{{BD}}$

Prove: $ \displaystyle \vartriangle ABD$ is isosceles

Solution:

Proof:

Statements	Reasons
1. $ \displaystyle \overline{{AC}}$ is the bisector of $ \overline{{BD}}$	1. Given
2. $ \displaystyle \overline{{BC}}\cong \overline{{CD}}$	2. Def of Bisector
3. $ \displaystyle \angle \,3$ and $ \displaystyle \angle \,4$ are right angles	3. $ \displaystyle \bot \,\,\Rightarrow \,\,\text{Right}\,\angle \text{ }\!\!’\!\!\text{ s}$
4. $ \displaystyle \angle \,3\cong \angle \,4$	4. Right Angles are $ \cong $
5. $ \displaystyle \overline{{AC}}\cong \overline{{AC}}$	5. Reflexive
6. $ \displaystyle \vartriangle ACB\cong \vartriangle ACD$	6. SAS
7. $ \displaystyle \overline{{AB}}\cong \overline{{AD}}$	5. CPCTC
8. $ \displaystyle \vartriangle ABD$ is isosceles	5. Def of Isos

Problem:

Using an Indirect Proof, prove the following:

Given: $ \displaystyle \angle \,1\cong \angle \,2$ and $ \displaystyle \overline{{BC}}$ is not congruent to $ \displaystyle \overline{{DC}}$

Prove: $ \displaystyle \overline{{AB}}$ is not congruent to $ \displaystyle \overline{{AD}}$

Solution:

Assume that $ \displaystyle \overline{{AB}}\cong \overline{{AD}}$, by assuming the conclusion is false.
$ \displaystyle \overline{{AC}}\cong \overline{{AC}}$ by reflexive.
$ \displaystyle \angle \,1\cong \angle \,2$ is given.
$ \displaystyle \vartriangle ACB\cong \vartriangle ACD$ by SAS.
$ \displaystyle \overline{{BC}}\cong \overline{{DC}}$ by CPCTC.
This contradicts the given information, so the assumption in step 1 must
be false.
Therefore, $ \displaystyle \overline{{AB}}$ is not congruent to $ \displaystyle \overline{{AD}}$

Trust me; these take practice!!

On to Polygons!