⌕
// STUDY REFERENCE

The full Geometry curriculum, distilled.

Ten chapters, every core theorem and formula, rendered with KaTeX. Built as a clean, dark-mode reference for studying or revisiting any topic on demand.

01·Foundations02·Reasoning03·Parallel Lines04·Congruence05·Similarity06·Right Triangles07·Quadrilaterals08·Circles09·Area10·Solids
01

Foundations of Geometry

Points, Lines, and Planes

A point marks a location and has no size. A line is a straight one-dimensional figure extending infinitely in both directions. A plane is a flat two-dimensional surface extending infinitely. Two points determine a line; three non-collinear points determine a plane.

ABCline ABplane
▤ Key Formulas
AB↔ denotes the line through A and B\overleftrightarrow{AB} \text{ denotes the line through } A \text{ and } B
AB‾ denotes the segment from A to B\overline{AB} \text{ denotes the segment from } A \text{ to } B
AB→ denotes the ray from A through B\overrightarrow{AB} \text{ denotes the ray from } A \text{ through } B
◇ Example 1
Problem: Name the plane containing A(1,0,0), B(0,1,0), C(0,0,1).
  1. The three points are non-collinear.
  2. They determine exactly one plane, often written as plane ABC.

Distance & Midpoint

The distance between two points in the plane is found with the Pythagorean-based distance formula. The midpoint is the average of the coordinates.

ABMdΔxΔy
▤ Key Formulas
d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
M=(x1+x22,y1+y22)M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)
◇ Example 1
Problem: Distance and midpoint between A(1, 2) and B(7, 10).
  1. d=(7−1)2+(10−2)2=36+64=10d = \sqrt{(7-1)^2 + (10-2)^2} = \sqrt{36 + 64} = 10
  2. M = ((1+7)/2, (2+10)/2) = (4, 6)

Angle Basics

An angle is formed by two rays with a common endpoint (vertex). Angles are classified by measure: acute (<90°), right (=90°), obtuse (90°<θ<180°), straight (=180°). Complementary angles sum to 90°; supplementary angles sum to 180°.

acuterightobtusestraight<90°=90°90°–180°=180°
▤ Key Formulas
Complementary: α+β=90∘\text{Complementary: } \alpha + \beta = 90^\circ
Supplementary: α+β=180∘\text{Supplementary: } \alpha + \beta = 180^\circ
◇ Example 1
Problem: An angle is 34°. Find its complement and supplement.
  1. Complement: 90 - 34 = 56°
  2. Supplement: 180 - 34 = 146°
Interactive · Distance Between Two Points

Compute the distance between (x₁, y₁) and (x₂, y₂) with the Pythagorean-based distance formula.

Flagship formula
d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
Try it — plug in numbers
Substituted
d=(7−1)2+(10−2)2d = \sqrt{(7 - 1)^2 + (10 - 2)^2}
Result
d=10d = 10
02

Reasoning & Proof

Inductive vs. Deductive Reasoning

Inductive reasoning uses patterns to form a conjecture. Deductive reasoning uses accepted facts and logic to reach a guaranteed conclusion. A single counterexample disproves a conjecture.

◇ Example 1
Problem: Conjecture: the sum of two odd numbers is odd. True?
  1. Counterexample: 3 + 5 = 8, which is even.
  2. The conjecture is false.

Conditional Statements

A conditional has the form ‘if p, then q’. Its converse swaps p and q. The inverse negates both. The contrapositive negates and swaps. A conditional and its contrapositive are logically equivalent.

pqif p, then q ( p ⊂ q )
▤ Key Formulas
Conditional: p⇒q\text{Conditional: } p \Rightarrow q
Converse: q⇒p\text{Converse: } q \Rightarrow p
Contrapositive: ¬q⇒¬p\text{Contrapositive: } \lnot q \Rightarrow \lnot p
◇ Example 1
Problem: Write the contrapositive of ‘If it rains, the ground is wet.’
  1. Negate and swap.
  2. ‘If the ground is not wet, then it is not raining.’

Two-Column Proofs

A two-column proof lists statements on the left and reasons on the right. Reasons include given information, definitions, postulates, and previously proven theorems. Common tools: the reflexive, symmetric, and transitive properties of equality.

◇ Example 1
Problem: Given AB = CD and CD = EF, prove AB = EF.
  1. AB = CD (given)
  2. CD = EF (given)
  3. AB = EF (transitive property of equality)
Interactive · Triangle Angle Sum

Given two angles of a triangle, find the third using the 180° sum.

Flagship formula
m∠A+m∠B+m∠C=180∘m\angle A + m\angle B + m\angle C = 180^\circ
Try it — plug in numbers
Substituted
m∠C=180−55−65m\angle C = 180 - 55 - 65
Result
m∠C=60∘m\angle C = 60^\circ
03

Parallel & Perpendicular Lines

Angles Formed by a Transversal

When a transversal cuts two parallel lines, corresponding angles are congruent, alternate interior angles are congruent, alternate exterior angles are congruent, and consecutive (same-side) interior angles are supplementary.

12345678ℓmt
▤ Key Formulas
Corresponding: ∠1≅∠5\text{Corresponding: } \angle 1 \cong \angle 5
Alt. interior: ∠3≅∠6\text{Alt. interior: } \angle 3 \cong \angle 6
Co-interior: m∠3+m∠5=180∘\text{Co-interior: } m\angle 3 + m\angle 5 = 180^\circ
◇ Example 1
Problem: Lines ℓ ∥ m cut by transversal t. ∠1 = 72°. Find its alternate interior angle.
  1. Alternate interior angles are congruent when lines are parallel.
  2. The alternate interior angle also measures 72°.

Slopes of Parallel & Perpendicular Lines

Two non-vertical lines are parallel exactly when their slopes are equal. They are perpendicular exactly when the product of their slopes is −1 (i.e., slopes are negative reciprocals).

m = ½m = ½m = −2m₁ · m₂ = −1
▤ Key Formulas
Parallel: m1=m2\text{Parallel: } m_1 = m_2
Perpendicular: m1⋅m2=−1\text{Perpendicular: } m_1 \cdot m_2 = -1
◇ Example 1
Problem: Line through (0, 1) perpendicular to y = 2x + 3.
  1. Slope of given line: 2. Perpendicular slope: -1/2.
  2. y−1=−12(x−0)⇒y=−12x+1y - 1 = -\tfrac{1}{2}(x - 0) \Rightarrow y = -\tfrac{1}{2}x + 1

Constructing Parallels & Perpendiculars

Using compass and straightedge, a line parallel to a given line through an external point can be constructed by copying an angle. A perpendicular through a point uses two arcs of equal radius to bisect a segment.

Pℓ∥ ℓ⟂ ℓ
Interactive · Perpendicular Slope

Enter a slope; get the perpendicular (negative reciprocal) slope.

Flagship formula
m1⋅m2=−1m_1 \cdot m_2 = -1
Try it — plug in numbers
Substituted
m2=−12m_2 = -\dfrac{1}{2}
Result
m2=−0.5m_2 = -0.5
04

Triangles & Congruence

Triangle Angle Sum & Exterior Angle

The three interior angles of any triangle sum to 180°. An exterior angle equals the sum of its two remote interior angles.

αγα+γABC
▤ Key Formulas
m∠A+m∠B+m∠C=180∘m\angle A + m\angle B + m\angle C = 180^\circ
m∠ext=m∠1+m∠2m\angle \text{ext} = m\angle_1 + m\angle_2
◇ Example 1
Problem: In △ABC, m∠A = 55°, m∠B = 65°. Find m∠C.
  1. m∠C = 180 - 55 - 65 = 60°

Congruence Postulates

Two triangles are congruent when a matching set of parts matches: SSS, SAS, ASA, AAS, and HL (right triangles only). SSA is NOT a valid congruence criterion.

≅SSS · SAS · ASA · AAS · HL
▤ Key Formulas
SSS, SAS, ASA, AAS, HL\text{SSS, SAS, ASA, AAS, HL}
◇ Example 1
Problem: △ABC and △DEF: AB = DE, ∠A = ∠D, AC = DF. Which postulate?
  1. Two sides and the included angle are congruent.
  2. By SAS, △ABC ≅ △DEF.

CPCTC

Once two triangles are proven congruent, ‘Corresponding Parts of Congruent Triangles are Congruent’ can be cited to conclude any pair of corresponding sides or angles are equal.

ABCDEF≅△ABC ≅ △DEF ⇒ ∠A ≅ ∠D, AB ≅ DE …
◇ Example 1
Problem: △ABC ≅ △XYZ. What is m∠B if m∠Y = 47°?
  1. ∠B corresponds to ∠Y.
  2. By CPCTC, m∠B = 47°.

Isosceles & Equilateral Triangles

In an isosceles triangle, the base angles (opposite the congruent sides) are congruent. An equilateral triangle has three 60° angles.

ABCM
▤ Key Formulas
Isosceles: AB=AC⇒∠B≅∠C\text{Isosceles: } AB = AC \Rightarrow \angle B \cong \angle C
Equilateral: each angle=60∘\text{Equilateral: each angle} = 60^\circ
Interactive · SAS Congruence Check

Test two triangles for congruence by Side–Angle–Side.

Flagship formula
SAS: AB=DE, ∠A=∠D, AC=DF\text{SAS: } AB = DE,\ \angle A = \angle D,\ AC = DF
Try it — plug in numbers
Substituted
AB=6, DE=6, ∠A=50∘, ∠D=50∘, AC=8, DF=8AB=6,\ DE=6,\ \angle A=50^\circ,\ \angle D=50^\circ,\ AC=8,\ DF=8
Result
△ABC≅△DEF\triangle ABC \cong \triangle DEF
05

Similarity

Similar Polygons

Two polygons are similar when corresponding angles are congruent and corresponding sides are proportional. The common ratio is the scale factor k.

akabkbscale factor k
▤ Key Formulas
ABA′B′=BCB′C′=CAC′A′=k\dfrac{AB}{A'B'} = \dfrac{BC}{B'C'} = \dfrac{CA}{C'A'} = k

Triangle Similarity: AA, SAS~, SSS~

Two triangles are similar when: two pairs of angles are congruent (AA); an angle is congruent and its including sides are proportional (SAS~); or all three sides are proportional (SSS~).

AA ⇒ similar
◇ Example 1
Problem: △ABC ~ △DEF with AB = 6, DE = 9, BC = 8. Find EF.
  1. Scale factor from △ABC to △DEF: 9/6 = 3/2.
  2. EF = 8 · (3/2) = 12

Side-Splitter & Angle Bisector Theorems

A line parallel to one side of a triangle divides the other two sides proportionally. An angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides.

ABCDExyuvx / y = u / v
▤ Key Formulas
ADDB=AEEC(DE∥BC)\dfrac{AD}{DB} = \dfrac{AE}{EC} \quad (DE \parallel BC)
BDDC=ABAC(AD bisects ∠A)\dfrac{BD}{DC} = \dfrac{AB}{AC} \quad (\text{AD bisects } \angle A)

Ratios of Perimeters, Areas, Volumes

If two similar figures have scale factor k, their perimeters are in ratio k, their areas in ratio k², and their volumes in ratio k³.

akaperimeter · karea · k²volume · k³
▤ Key Formulas
Perimeters: k\text{Perimeters: } k
Areas: k2\text{Areas: } k^2
Volumes: k3\text{Volumes: } k^3
Interactive · Similarity Scale Factor

Compute the scale factor between similar triangles and map a side length.

Flagship formula
k=A′B′AB,E′F′=k⋅EFk = \dfrac{A'B'}{AB},\quad E'F' = k \cdot EF
Try it — plug in numbers
Substituted
k=9/6=1.5, E′F′=8⋅1.5k = 9/6 = 1.5,\ E'F' = 8 \cdot 1.5
Result
E′F′=12E'F' = 12
06

Right Triangles & Trigonometry

Pythagorean Theorem & Converse

In any right triangle with legs a, b and hypotenuse c, a² + b² = c². Its converse: if a² + b² = c² for a triangle's sides, then the triangle is right.

a²b²c²abc
▤ Key Formulas
a2+b2=c2a^2 + b^2 = c^2
◇ Example 1
Problem: Find c if a = 5, b = 12.
  1. c² = 25 + 144 = 169
  2. c = 13

Special Right Triangles

A 45-45-90 triangle has legs equal and hypotenuse leg·√2. A 30-60-90 triangle has sides in the ratio 1 : √3 : 2.

xxx√245°45°x√3x2x30°60°
▤ Key Formulas
45-45-90: 1:1:245\text{-}45\text{-}90:\ 1 : 1 : \sqrt{2}
30-60-90: 1:3:230\text{-}60\text{-}90:\ 1 : \sqrt{3} : 2
◇ Example 1
Problem: Legs of a 45-45-90 have length 7. Hypotenuse?
  1. Hypotenuse=72\text{Hypotenuse} = 7\sqrt{2}

Trigonometric Ratios (SOH-CAH-TOA)

For an acute angle θ in a right triangle: sine is opposite/hypotenuse, cosine is adjacent/hypotenuse, tangent is opposite/adjacent.

θadjacentoppositehypotenusesin θ = opp / hyp
▤ Key Formulas
sin⁡θ=opphyp\sin\theta = \dfrac{\text{opp}}{\text{hyp}}
cos⁡θ=adjhyp\cos\theta = \dfrac{\text{adj}}{\text{hyp}}
tan⁡θ=oppadj\tan\theta = \dfrac{\text{opp}}{\text{adj}}
◇ Example 1
Problem: θ has opp = 3, adj = 4, hyp = 5. Find sin θ, cos θ, tan θ.
  1. sin⁡θ=3/5\sin\theta = 3/5
  2. cos⁡θ=4/5\cos\theta = 4/5
  3. tan⁡θ=3/4\tan\theta = 3/4

Angles of Elevation & Depression

An angle of elevation is measured up from the horizontal; an angle of depression, down. In a diagram the two are alternate interior angles and therefore equal.

elevationdepressionobserver
◇ Example 1
Problem: Sighting the top of a 40 ft flagpole from 30 ft away, angle of elevation?
  1. tan⁡θ=40/30=4/3\tan\theta = 40/30 = 4/3
  2. θ=arctan⁡(4/3)≈53.13∘\theta = \arctan(4/3) \approx 53.13^\circ
Interactive · Right-Triangle Trig

Enter the legs; get the hypotenuse and every basic trig ratio.

Flagship formula
a2+b2=c2,  tan⁡θ=aba^2 + b^2 = c^2,\ \ \tan\theta = \tfrac{a}{b}
Try it — plug in numbers
Substituted
c=32+42, tan⁡θ=3/4c = \sqrt{3^2 + 4^2},\ \tan\theta = 3/4
Result
c=5, θ≈36.87∘, sin⁡θ=0.6, cos⁡θ=0.8, tan⁡θ=0.75c = 5,\ \theta \approx 36.87^\circ,\ \sin\theta = 0.6,\ \cos\theta = 0.8,\ \tan\theta = 0.75
07

Quadrilaterals & Polygons

Polygon Angle Sums

The sum of interior angles of an n-gon is (n − 2)·180°. The sum of exterior angles (one per vertex) is always 360°.

(n − 2) · 180° = 4 · 180°
▤ Key Formulas
Sint=(n−2)⋅180∘S_{\text{int}} = (n - 2)\cdot 180^\circ
Sext=360∘S_{\text{ext}} = 360^\circ
◇ Example 1
Problem: Sum of interior angles of a hexagon.
  1. (6 - 2) · 180 = 720°

Parallelograms

In a parallelogram both pairs of opposite sides are parallel and congruent, opposite angles are congruent, consecutive angles are supplementary, and the diagonals bisect each other.

ABCD

Rectangles, Rhombuses, Squares

A rectangle is a parallelogram with four right angles (diagonals are congruent). A rhombus is a parallelogram with four congruent sides (diagonals are perpendicular and bisect the vertex angles). A square is both — every property of rectangles and rhombuses holds.

rectanglerhombussquare= diagonals⟂ diagonalsboth

Trapezoids & Kites

A trapezoid has exactly one pair of parallel sides (the bases). The midsegment is parallel to the bases and equal to their average. A kite has two pairs of consecutive congruent sides; its diagonals are perpendicular.

b₁b₂midsegment
▤ Key Formulas
Midsegment: m=b1+b22\text{Midsegment: } m = \dfrac{b_1 + b_2}{2}
Interactive · Polygon Angle Sum

For an n-gon, compute the total interior angle sum and each regular angle.

Flagship formula
S=(n−2)⋅180∘S = (n - 2)\cdot 180^\circ
Try it — plug in numbers
Substituted
S=(6−2)⋅180S = (6 - 2)\cdot 180
Result
S=720∘, each (regular)=120∘S = 720^\circ,\ \text{each (regular)} = 120^\circ
08

Circles

Circumference & Arc Length

The circumference of a circle of radius r is 2πr. An arc's length is a fraction of that, proportional to its central angle.

rθss = r θ
▤ Key Formulas
C=2πrC = 2\pi r
Arc length=θ360∘⋅2πr\text{Arc length} = \dfrac{\theta}{360^\circ} \cdot 2\pi r
◇ Example 1
Problem: Arc length for θ = 90° on a circle of radius 6.
  1. (90/360) · 2π · 6 = (1/4) · 12π = 3π

Central & Inscribed Angles

A central angle equals its intercepted arc. An inscribed angle equals half its intercepted arc. Inscribed angles that intercept the same arc are congruent, and an inscribed angle in a semicircle is a right angle.

2θθABCO
▤ Key Formulas
Central: m∠=mAB⌢\text{Central: } m\angle = m\overset{\frown}{AB}
Inscribed: m∠=12 mAB⌢\text{Inscribed: } m\angle = \tfrac{1}{2}\, m\overset{\frown}{AB}

Chords, Tangents, and Secants

A tangent is perpendicular to the radius drawn to the point of tangency. Two tangents from an external point are congruent. Products of chord/secant segments follow the intersecting-chords and secant-tangent power rules.

tangentrchordsecant
▤ Key Formulas
Chords: AE⋅EB=CE⋅ED\text{Chords: } AE \cdot EB = CE \cdot ED
Secant-tangent: t2=a(a+b)\text{Secant-tangent: } t^2 = a(a + b)

Equation of a Circle

A circle with center (h, k) and radius r has the standard equation shown. Complete the square to convert general form to standard form.

(h, k)(x, y)rhk(x−h)² + (y−k)² = r²
▤ Key Formulas
(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2
◇ Example 1
Problem: Center and radius of x² + y² − 4x + 6y = 12.
  1. (x−2)2−4+(y+3)2−9=12(x-2)^2 - 4 + (y+3)^2 - 9 = 12
  2. (x−2)2+(y+3)2=25(x-2)^2 + (y+3)^2 = 25
  3. Center (2, -3), radius 5
Interactive · Point-on-Circle Checker

Given a circle (h, k, r), classify a point as inside, on, or outside.

Flagship formula
(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2
Try it — plug in numbers
Substituted
(5−2)2+(1−−3)2  ?  52(5 - 2)^2 + (1 - -3)^2 \;?\; 5^2
Result
25  ?  25⇒on the circle25 \;?\; 25 \Rightarrow \text{on the circle}
09

Area & Perimeter

Triangles & Parallelograms

The area of a triangle is one-half base times height. A parallelogram's area is base times height (the perpendicular distance between the parallel sides).

bhA = b · h
▤ Key Formulas
A△=12 b hA_{\triangle} = \tfrac{1}{2}\,b\,h
A□=b hA_{\square} = b\,h
◇ Example 1
Problem: Triangle with base 10 and height 6.
  1. A = (1/2)(10)(6) = 30

Trapezoids & Regular Polygons

A trapezoid's area uses the average of the bases times the height. A regular polygon's area uses one-half its apothem times its perimeter.

b₁b₂has
▤ Key Formulas
Atrap=12(b1+b2)hA_{\text{trap}} = \tfrac{1}{2}(b_1 + b_2)h
Areg=12 a PA_{\text{reg}} = \tfrac{1}{2}\,a\,P

Circles & Sectors

A circle of radius r has area πr². A sector is a fraction of that area proportional to its central angle.

θrA = ½ r² θ
▤ Key Formulas
A∘=πr2A_{\circ} = \pi r^2
Asector=θ360∘ πr2A_{\text{sector}} = \dfrac{\theta}{360^\circ}\,\pi r^2
◇ Example 1
Problem: Sector area for θ = 60°, r = 6.
  1. (60/360) · π · 36 = 6π

Composite Figures

For irregular shapes, decompose into familiar pieces (triangles, rectangles, circles), add or subtract areas as needed, and combine.

rectanglesemicircletotal = rectangle + semicircle
Interactive · Sector Area

Compute the area of a circular sector from θ and r.

Flagship formula
Asector=θ360∘ πr2A_{\text{sector}} = \dfrac{\theta}{360^\circ}\,\pi r^2
Try it — plug in numbers
Substituted
A=60360⋅π(6)2A = \dfrac{60}{360}\cdot \pi (6)^2
Result
A≈18.8496A \approx 18.8496
10

Surface Area & Volume

Prisms & Cylinders

For any right prism or cylinder, the volume is base area times height. The lateral surface area is the perimeter (or circumference) of the base times the height.

lhwrh
▤ Key Formulas
Vprism=B hV_{\text{prism}} = B\,h
Vcyl=πr2hV_{\text{cyl}} = \pi r^2 h
SAcyl=2πr2+2πrhSA_{\text{cyl}} = 2\pi r^2 + 2\pi r h
◇ Example 1
Problem: Volume of a cylinder with r = 3, h = 10.
  1. V = π(9)(10) = 90π

Pyramids & Cones

A pyramid or cone has one-third the volume of the prism or cylinder with the same base and height. Slant height ℓ is used for lateral surface area of a cone.

hlhrl
▤ Key Formulas
Vpyr=13BhV_{\text{pyr}} = \tfrac{1}{3} B h
Vcone=13πr2hV_{\text{cone}} = \tfrac{1}{3}\pi r^2 h
SAcone=πr2+πrℓSA_{\text{cone}} = \pi r^2 + \pi r \ell

Spheres

A sphere of radius r has surface area 4πr² and volume (4/3)πr³.

rV = 4/3 π r³ S = 4π r²
▤ Key Formulas
SA⊙=4πr2SA_{\odot} = 4\pi r^2
V⊙=43πr3V_{\odot} = \tfrac{4}{3}\pi r^3
◇ Example 1
Problem: Volume of a sphere with r = 3.
  1. V = (4/3)π(27) = 36π

Similar Solids

If two similar solids have linear scale factor k, their surface areas are in ratio k² and their volumes in ratio k³.

a2asides 1 : 2areas 1 : 4volumes 1 : 8
▤ Key Formulas
SA1SA2=k2\dfrac{SA_1}{SA_2} = k^2
V1V2=k3\dfrac{V_1}{V_2} = k^3
Interactive · Sphere Volume & Surface Area

Compute V = (4/3)πr³ and SA = 4πr² for a sphere.

Flagship formula
V=43πr3,  SA=4πr2V = \tfrac{4}{3}\pi r^3,\ \ SA = 4\pi r^2
Try it — plug in numbers
Substituted
V=43π(3)3, SA=4π(3)2V = \tfrac{4}{3}\pi (3)^3,\ SA = 4\pi (3)^2
Result
V≈113.0973, SA≈113.0973V \approx 113.0973,\ SA \approx 113.0973