Comprehensive Guide to Geometry Axioms, Theorems, and Formulas

Axioms of Inequality and Algebraic Relationships

The principles of inequality define how relative values change or remain consistent under specific operations. If b>ab > a, then it follows that c+b>c+ac + b > c + a, cb>cac - b > c - a, and c×b>c×ac \times b > c \times a. Conversely, if b<ab < a, then d+b<c+ad + b < c + a. In cases where an addition results in a sum, such as if c+b=ac + b = a, it can be concluded that c<ac < a and b<ab < a is false, rather the transcript states c>ac > a and b>ab > a is not possible if they are parts of the whole, implying aa is the larger sum; specifically, the rule notes if c+b=ac + b = a, then c<ac < a and b<ab < a (assuming positive values). Additional logical deductions for angles include: if m(ABD)>m(XYL)m(\angle ABD) > m(\angle XYL) and m(DBC)<m(LYZ)m(\angle DBC) < m(\angle LYZ), various comparisons can be made regarding the composite angles.

Exterior Angles and Triangle Inequalities

The Exterior Angle Theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. Formally, for a triangle ABCABC with an exterior angle at CC labeled as ACDACD, the relationship is m(ACD)=m(A)+m(B)m(\angle ACD) = m(\angle A) + m(\angle B). Because the measure of the exterior angle is the sum of these two interior angles, it inherently follows that the exterior angle is greater than either individual interior angle: m(ACD)>m(A)m(\angle ACD) > m(\angle A) and m(ACD)>m(B)m(\angle ACD) > m(\angle B). Furthermore, in any triangle, the side opposite a greater angle is longer than the side opposite a smaller angle. This is expressed in Rule 1: If AC>ABAC > AB, then m(B)>m(C)m(\angle B) > m(\angle C). Conversely, if m(B)>m(C)m(\angle B) > m(\angle C), then AC>ABAC > AB. A specific note for a single triangle emphasizes that the greatest side is always opposite the greatest angle; for instance, if BB is the greatest angle (such as an obtuse angle), then ACAC is the longest side, leading to the inequalities AB<ACAB < AC and BC<ACBC < AC.

The Triangle Inequality Theorem

The triangle inequality theorem establishes the necessary relationship between the lengths of the three sides of a triangle (ABAB, BCBC, and ACAC). It states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This can be summarized by Ruler 2: BCAC<AB<BC+AC|BC - AC| < AB < BC + AC. This inequality shows that the length of side ABAB must be strictly greater than the absolute difference of the other two sides and strictly less than their sum. If the sum of two sides is equal to the third side (AC=BC+ABAC = BC + AB), then the points AA, BB, and CC are collinear. If the sum of two sides is less than the third side (AC>BC+ABAC > BC + AB), the points AA, BB, and CC are considered vertices of a triangle only if the inequality holds; however, specifically, the notes state if AC>BC+ABAC > BC + AB, they are vertices.

Geometric Projections and Theorems of Pythagoras and Euclid

Projections describe the mapping of points or segments onto a line. The projection of a point CC onto a line LL is a point, and the length of the projection of a segment is always less than or equal to the length of the segment itself (length of projectionlength of segment\text{length of projection} \le \text{length of segment}). If a segment is perpendicular to the line (ABLAB \perp L), the length of the projection is 00. If a segment is parallel to the line (ABLAB \parallel L), then the length of the projection is equal to the original segment (AB=ABA'B' = AB).

In right-angled triangle calculations, Pythagoras' theorem provides the relations: (AC)2=(AB)2+(BC)2(AC)^2 = (AB)^2 + (BC)^2, (AB)2=(AC)2(BC)2(AB)^2 = (AC)^2 - (BC)^2, and (BC)2=(AC)2(AB)2(BC)^2 = (AC)^2 - (AB)^2. Euclid's theorems provide additional relations for a right triangle with an altitude ADAD dropped from the right angle AA to the hypotenuse BCBC:

  1. (AB)2=BC×BD(AB)^2 = BC \times BD
  2. (AC)2=CB×CD(AC)^2 = CB \times CD
  3. (AD)2=DC×DB(AD)^2 = DC \times DB
  4. AD=AC×ABBCAD = \frac{AC \times AB}{BC}

Circle Geometry and Coordinate Calculations

A circle is defined by its center MM and radius rr (segments such as MAMA or MBMB). The diameter dd is defined as d=2rd = 2r. The circumference of a circle is calculated as C=2πrC = 2\pi r or C=πdC = \pi d. The area of a circle is given by Area=πr2Area = \pi r^2, where π\pi is approximately 227\frac{22}{7} or 3.143.14.

In coordinate geometry, the location of a point (x,y)(x, y) relative to the axes is determined by absolute values: the distance to the x-axis is y|y| (e.g., for point (4,3)(4, -3), the distance is 3=3|-3| = 3 units), and the distance to the y-axis is x|x|. The distance dd between two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) is calculated using the formula: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. The distance from a point A(x,y)A(x, y) to the origin B(0,0)B(0, 0) is d=x2+y2d = \sqrt{x^2 + y^2}. The slope mm of a line passing through these points is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. If a line is parallel to the x-axis, its slope is 00. If a line is parallel to the y-axis, its slope is undefined. For points to be collinear, the slope of ABAB must equal the slope of BCBC. To prove points AA, BB, and CC lie on the same circle with center MM, one must prove MA=MB=MCMA = MB = MC.

Classification of Triangles by Sides and Angles

Triangles are classified by side lengths: an equilateral triangle has AB=BC=ACAB = BC = AC, and an isosceles triangle has at least two equal sides (e.g., AB=ACAB = AC). To determine if a triangle is acute, right, or obtuse at vertex BB, compare the square of the longest side (assume ACAC) to the sum of the squares of the other two sides:

  1. If (AC)2<(AB)2+(BC)2(AC)^2 < (AB)^2 + (BC)^2, the triangle is acute.
  2. If (AC)2=(AB)2+(BC)2(AC)^2 = (AB)^2 + (BC)^2, the triangle is right-angled at BB.
  3. If (AC)2>(AB)2+(BC)2(AC)^2 > (AB)^2 + (BC)^2, the triangle is obtuse at BB.

General area and perimeter formulas for triangles include: Area=12×base×heightArea = \frac{1}{2} \times \text{base} \times \text{height} and Perimeter=Sum of all sidesPerimeter = \text{Sum of all sides}. For a square with side SS, Area=S2Area = S^2 and Perimeter=4SPerimeter = 4S. For a rectangle with length LL and width WW, Area=L×WArea = L \times W and Perimeter=2×(L+W)Perimeter = 2 \times (L + W).

Properties and Volume of Geometric Solids

The volume and surface area of a Right Circular Cylinder are determined by the radius rr and height hh:

  • Volume=πr2hVolume = \pi r^2 h (Base Area ×\times Height)
  • LateralArea=2πrhLateral Area = 2\pi rh (Base Circumference ×\times Height)
  • TotalArea=2πrh+2πr2Total Area = 2\pi rh + 2\pi r^2 (Lateral Area + Double Base Area), which simplifies to 2πr(r+h)2\pi r(r + h).

For a Right Prism (including Triangular and Quadrilateral Prisms):

  • Volume=base area×heightVolume = \text{base area} \times \text{height}
  • LateralArea=base perimeter×heightLateral Area = \text{base perimeter} \times \text{height}
  • TotalArea=lateral area+(2×base area)Total Area = \text{lateral area} + (2 \times \text{base area}).

Identifying Specific Quadrilaterals

To prove a quadrilateral ABCDABCD is a specific shape, certain conditions must be met:

  1. Parallelogram: Prove opposite sides are equal (AD=BCAD = BC and CD=ABCD = AB).
  2. Rhombus: Prove all sides are equal (AD=CD=BC=ABAD = CD = BC = AB).
  3. Rectangle: Prove it is a parallelogram (AD=BCAD = BC and CD=ABCD = AB) and the diagonals are equal (AC=BDAC = BD).
  4. Square: Prove all sides are equal (AD=CD=BC=ABAD = CD = BC = AB) and the diagonals are equal (AC=BDAC = BD).