Properties, Postulates, and Definitions in High School Geometry

Reflexive, Substitution, and Transitive Properties

Formal geometric proofs are a primary focus in geometry. Before engaging in complex proofs, specific properties and postulates must be understood.

Reflexive Property

• The reflexive property states that an angle or segment is always congruent to or equal to itself.
• This is used when one angle or segment is shared by two different scenarios or shapes within a single diagram.
• In a diagram where points $B$ and $C$ form a segment $BC$ on a larger line segment $AD$, $BC$ is a part of segment $AC$ and also a part of segment $BD$. The reflexive property asserts that the $BC$ within $AC$ is congruent to the $BC$ within $BD$ because they are the identical segment. Therefore: \text{BC} \bcong \text{BC}.
• This property applies to angles as well. In a diagram of $\triangle ABC$ and $\triangle DBE$ where \bangle B is a shared angle, the reflexive property states that the \bangle B in $\triangle ABC$ is congruent to the \bangle B in $\triangle DBE$. Therefore: \bangle B \bcong \bangle B.

Substitution Property

• The substitution property states that if $a = b$ in a question, then any instance of the variable $b$ in that question can be rewritten as $a$.

Transitive Property

• The transitive property states that if $a = b$ and $b = c$, then it follows that $a = c$.
• Example: If Lorraine is in Jerry's English class and Jerry is in Madeline's English class, then by the transitive property, Lorraine must be in Madeline's English class.

Addition and Subtraction Postulates

Postulates are used when moving between larger segments/angles and their smaller constituent parts.

Definitions

• Addition Postulate: If two congruent segments or angles are added to two segments or angles that are already congruent, the resulting sums will also be congruent.
• Subtraction Postulate: If two congruent segments or angles are subtracted from two segments or angles that are already congruent, the resulting differences will also be congruent.
• Determining Use: Subtraction is used when the measurement begins with larger angles and ends with smaller ones. Addition is used when the measurement begins with smaller parts and ends with a larger whole.

Segment Example using Addition

In a segment labeled $M-A-T-H$, if it is given that \text{MA} \bcong \text{TH}, the addition postulate can prove \text{MT} \bcong \text{HA}.

1. By the reflexive property, the shared middle segment is congruent to itself: \text{AT} \bcong \text{AT}.
2. The diagram shows that the whole segments are sums: $\text{MA} + \text{AT} = \text{MT}$ and $\text{HT} + \text{AT} = \text{HA}$.
3. Because we added the same congruent piece ($\text{AT}$) to the already congruent pieces ($\text{MA}$ and $\text{TH}$), then \text{MT} \bcong \text{HA}.

Geometrical Proof Example using Subtraction

In a diagram of segment $ABC$ with angles \bangle ABE and \bangle CBD, given that \bangle ABE \bcong \bangle CBD, we prove that \bangle ABD \bcong \bangle CBE.

• Context: \bangle ABE is the sum of \bangle ABD and \bangle DBE. \bangle CBD is the sum of \bangle CBE and \bangle DBE. Both large angles share the interior angle \bangle DBE.
• Process: To prove the outer small angles congruent, subtract the shared piece (\bangle DBE).

Formal Proof Table

• Calculation Logic:
  • \bangle ABE - \bangle DBE = \bangle ABD
  • \bangle CBD - \bangle DBE = \bangle CBE

Angle Relationships and Definitions

Complementary and Supplementary Angles

• Complementary Angles: A pair of angles whose sum is exactly $90^{\circ}$. One angle is called the complement of the other.
• Supplementary Angles: A pair of angles whose sum is exactly $180^{\circ}$. One angle is called the supplement of the other.
• Linear Pairs: Two adjacent angles that form a straight line. Because a straight angle equals $180^{\circ}$, a linear pair consists of supplementary angles.
• Adjacent Angles: Angles that sit by each other, sharing a side and a vertex without overlapping.

Angle Bisectors

• An angle bisector is a line or ray that divides an angle into two congruent smaller angles.
• If a large angle is bisected, the two smaller parts added together equal the whole.

Intersection and Perpendicularity

Vertical Angles

• When two lines intersect, they form vertical angles that are congruent to one another.

Perpendicular Lines ($\perp$)

• Perpendicular lines are two lines that intersect to create right angles (represented by a box at the intersection point).
• All right angles equal $90^{\circ}$.
• All right angles are congruent to each other.
• If line AB \bperp CD at point $E$, then \bangle AEC, \bangle AED, \bangle BEC, and \bangle BED are all right angles and are all congruent: \bangle AEC \bcong \bangle AED \bcong \bangle BEC \bcong \bangle BED.

Perpendicular Bisector

• A perpendicular bisector does two things simultaneously: it intersects a segment at a $90^{\circ}$ angle (perpendicular) and divides that segment into two congruent parts (bisector).
• Example: If $AB$ is the perpendicular bisector of $CD$ at point $E$:
  1. $CD$ is bisected: \text{CE} \bcong \text{ED}.
  2. Right angles are formed at $E$: \bangle AEC, \bangle AED, \bangle BEC, and \bangle BED are all right angles.

Algebraic Applications in Geometry

Solving a Linear Pair

Given a diagram where \bangle AEC = 4x - 2 and \bangle CEB = x + 12 form a linear pair:

1. Step 1: Set the sum of the angles equal to $180^{\circ}$.     $4x - 2 + x + 12 = 180$
2. Step 2: Combine like terms.     $5x + 10 = 180$
3. Step 3: Subtract $10$ from both sides.     $5x = 170$
4. Step 4: Divide by $5$.     $x = 34$

Solving Complementary Angles Word Problem

Given two complementary angles where one angle is $9$ less than twice the other angle. Find the larger angle.

1. Step 1: Define variables.     Let the first angle be $x$.     Let the second angle be $2x - 9$.
2. Step 2: Set the sum equal to $90^{\circ}$.     $x + (2x - 9) = 90$
3. Step 3: Solve for $x$.     $3x - 9 = 90$$3x = 99$$x = 33$
4. Step 4: Find both measures.     Angle 1: $33^{\circ}$     Angle 2: $2(33) - 9 = 66 - 9 = 57^{\circ}$.
5. The larger angle is $57^{\circ}$.

Solving Angle Bisectors

Given $AE$ bisects \bangle DEF, where m\bangle DEF = 10x + 4 and m\bangle AED = 3x + 12. Find m\bangle DEF.

1. Step 1: Recognize that if \bangle DEF is bisected, \bangle AED \bcong \bangle AEF. Thus, m\bangle AEF is also $3x + 12$.
2. Step 2: Set the sum of the parts equal to the whole.     $(3x + 12) + (3x + 12) = 10x + 4$
3. Step 3: Solve for $x$.     $6x + 24 = 10x + 4$$24 = 4x + 4$$20 = 4x$$x = 5$
4. Step 4: Substitute $x$ into the whole angle expression.     m\bangle DEF = 10(5) + 4 = 54^{\circ}.

Solving Vertical Angles

Given intersecting lines $CT$ and $MP$ at point $A$, with m\bangle CAM = 12x + 8 and m\bangle PAT = 13x - 1:

1. Step 1: Set the vertical angles equal.     $12x + 8 = 13x - 1$
2. Step 2: Solve for $x$.     $8 = x - 1$$9 = x$

Solving Perpendicular Lines Algebraically

Given AM \bperp RS and m\bangle ATR = 4x - 5:

1. Step 1: Since the lines are perpendicular, the angle must equal $90^{\circ}$.     $4x - 5 = 90$
2. Step 2: Solve for $x$.     $4x = 95$$x = 23.75$