Classifying Triangles
Note: 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 that can be classified by angles, such as acute, obtuse, or right.
An acute triangle has three acute angles ($ displaystyle <90{}^circ $), an obtuse triangle has one obtuse angle ($ displaystyle >90{}^circ $), and a right triangle has one right angle ($90{}^circ $). Here are examples:
Triangles are also classified by sides, such as equilateral, isosceles, or scalene.
An equilateral triangle has three congruent sides, as well as three congruent angles, an isosceles has at least two sides that are congruent, with and congruent angles oppositive the sides, and a scalene triangle has no two sides that are congruent. Note that equilateral triangles are also isosceles. 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 know 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 (inside) angles that are not adjacent (remote interior angles):
$ displaystyle mangle 1 +mangle 2=mangle 3 $
Two corollaries come from this theorem; these make sense just by using some simple addition or subtraction:
- Acute angles of a right triangle are complementary.
- A triangle can have at most one right or obtuse angle.
There are a few Triangle Inequality Theorems that we’ll need to know about:
- 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 (three sides and three angles) that are congruent between the triangles; we can mark them to show which of the parts have the same measurement and thus are corresponding: 
Note that it’s very important to label the triangles correctly, so the angles of the first triangle line up with the respective congruent angles of the second triangle. Thus $ displaystyle vartriangle ABCcong vartriangle DEF$, but $ displaystyle vartriangle ABC$ is not necessarly $ cong$ to $ vartriangle DFE$, for example.
There are certain tests for congruency for triangles, so we don’t have to show that all corresponding parts, the three angles and three sides, are congruent. Here are 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 ABCcong 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 Bcong angle E,,,,overline{{BC}}cong overline{{EF}}$ $ displaystyle vartriangle ABCcong 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 Acong angle D,,,,angle Ccong angle F,,,,overline{{AB}}cong overline{{DE}}$
$ displaystyle vartriangle ABCcong vartriangle DEF$ |
| HL Postulate
(Hypotenuse-Leg Postulate) |
If the hypotenuse and another leg of a 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 ABCcong 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, after proving triangle congruence, that 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 ABCcong vartriangle DEF$, we have:
$ displaystyle overline{{AB}}cong overline{{DE}},,,,overline{{BC}}cong overline{{EF}},,,,overline{{AC}}cong overline{{DF}}$ and $ displaystyle angle Acong angle D,,,angle Bcong angle E,,,angle Ccong angle F$:
Just think of CPCTC as “picking apart” congruent triangles and lining up their respective corresponding parts!
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)
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: 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 mangle 1 +mangle 2=mangle 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 DABcong 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 ,1cong angle ,2$ | 4. $ displaystyle text{AIA}Rightarrow ,cong ,angle ,text{ }!!’!!text{ s}$ |
| 5. $ displaystyle vartriangle DABcong 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 ACBcong vartriangle ACD$ | 6. SAS |
| 7. $ displaystyle overline{{AB}}cong overline{{AD}}$ | 7. CPCTC |
Problem:
Given: $ displaystyle angle ABC$ and $ displaystyle angle BCD$ are right angles, $ displaystyle overline{{AC}}cong overline{{BD}}$
Prove: $ displaystyle angle ,1cong 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 ABCcong vartriangle DCB$ | 5. Hypotenuse-Leg (HL) |
| 6. $ displaystyle angle ,1cong angle ,2$ | 6. CPCTC |
Note that there may be easier ways to do this next proof, using some of the theorems we used:
Problem:
Given: $ displaystyle angle ,2cong angle ,3$, $ displaystyle angle ,7cong angle ,8$ 
Prove: $ displaystyle overline{{DF}}cong overline{{EF}}$
Solution:
Proof:
| Statements | Reasons |
| 1. $ displaystyle angle ,7cong 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 -> Suppl <‘s |
| 4.$ displaystyle mangle ,5 + mangle ,7=180$, $ displaystyle mangle ,6 + mangle ,8=180$ | 4. Def of Suppl |
| 5. $ displaystyle mangle ,5 =180- mangle ,7 $, $ displaystyle mangle ,6 =180- mangle ,8 $ | 5. Subtr = |
| 6. $ displaystyle mangle ,7 = mangle ,8$ | 6. Def of Congr |
| 7. $ displaystyle mangle ,5 =180- mangle ,7 $, $ displaystyle mangle ,6 =180- mangle ,7 $ | 7. Subst |
| 8. $ displaystyle mangle ,5 = mangle ,6$ | 8. Trans. |
| 9. $ displaystyle angle ,5cong angle ,6$ | 9. Def of Congr |
| 10. $ displaystyle overline{{AG}}cong overline{{AG}}$ | 10. Reflexive |
| 11. $ displaystyle vartriangle AGDcong vartriangle AGE$ | 9. ASA |