The logic of shape and space: proof, congruence, circles, trigonometry, and measurement in two and three dimensions.
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Lessons
Unit 1: Reasoning, Angles, and Triangles
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Lesson 1: Proofs and Logical Reasoning
Geometry is where mathematics learns to prove things. A proof is a chain of statements, each justified by a definition, postulate, or previously proven theorem, leading from what is given to what must be shown. The building blocks: postulates (accepted without proof, e.g. “through two points there is exactly one line”) and theorems (proven from postulates, e.g. “vertical angles are congruent”). Every step needs a reason — “it looks true in the diagram” is never a reason.
Two reasoning styles power geometry. Inductive reasoning spots patterns from examples (measuring many triangles' angles and conjecturing they sum to 180°); deductive reasoning derives certainties from general truths (since all triangles' angles sum to 180°, this triangle's do too). A single counterexample kills a conjecture but no number of examples proves one — that asymmetry is why proof matters.
The two-column proof organizes statements on the left and reasons on the right. Conditional statements (“if p, then q”) have a converse (“if q, then p”), inverse, and contrapositive — and only the contrapositive is logically equivalent to the original. “If it rains, the ground is wet” does not imply “if the ground is wet, it rained” (a sprinkler could be the cause). Indirect proof (proof by contradiction) assumes the opposite of what you want and derives an impossibility.
Worked example — proving vertical angles are congruent:
Given: lines AB and CD intersect at O, forming vertical angles ∠AOC and ∠BOD. Prove: ∠AOC ≅ ∠BOD.
∠AOC and ∠AOD form a linear pair, so m∠AOC + m∠AOD = 180° (Linear Pair Postulate).
∠BOD and ∠AOD also form a linear pair, so m∠BOD + m∠AOD = 180° (same reason).
Both sums equal 180°, so m∠AOC + m∠AOD = m∠BOD + m∠AOD (substitution).
Subtract m∠AOD from both sides: m∠AOC = m∠BOD (subtraction property of equality).
Equal measures mean congruent angles (definition of congruent angles). QED.
Try it yourself: Write the converse, inverse, and contrapositive of: “If a figure is a square, then it has four right angles.” Which are true? Then prove: if two angles are complementary to the same angle, they are congruent.
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Narrated audio version: Geometry: Shapes.
Lesson 2: Angles, Parallel Lines, and Transversals
When a transversal crosses two parallel lines, it creates eight angles with beautiful relationships. Corresponding angles sit in matching corners and are equal; alternate interior angles sit on opposite sides of the transversal between the lines and are equal; same-side interior angles are supplementary (sum 180°). Vertical angles (opposite each other at a crossing) are always equal, whether or not the lines are parallel.
These relationships let you solve for unknowns: if one angle is labeled (3x + 10)° and its corresponding angle is (5x − 20)°, set them equal: 3x + 10 = 5x − 20, so 2x = 30, x = 15, and both angles are 55°. The converse also holds: if corresponding angles are equal, the lines must be parallel — the test surveyors and carpenters use to check that walls are parallel.
Angle sums follow from the same ideas. A triangle's interior angles sum to 180° (draw a line through one vertex parallel to the opposite side and watch the alternate interior angles appear). A quadrilateral's sum is 360°, and an n-gon's interior sum is (n − 2) × 180°, because any polygon splits into n − 2 triangles from one vertex.
Worked example — two parallel lines cut by a transversal; one angle is 118°. Find the other seven:
Its vertical angle is also 118°. Its supplementary neighbors (on a straight line) are 180 − 118 = 62°.
Corresponding angles at the second crossing match: 118°, 118°, 62°, 62°.
So the eight angles are four 118° angles and four 62° angles.
Try it yourself: A transversal makes a 75° angle with one of two parallel lines. List the measures of all eight angles. A pentagon's interior angles sum to what? Answers: Four 75° angles and four 105° angles; (5 − 2) × 180 = 540°.
Download Worksheets
Print these worksheets for offline practice and homework. Each PDF includes an answer key on the last pages (for teacher/parent use).
Homework Assignment (PDF) — application-focused tasks, including an explain-in-your-own-words or project task.
Key vocabulary:Transversal — a line crossing two or more lines; corresponding angles — matching-corner angles, equal when lines are parallel; supplementary — two angles summing to 180°.
Lesson 3: Triangles and Congruence
Two figures are congruent if one can be moved (by rotations, reflections, translations) to coincide exactly with the other — same shape, same size. For triangles, proving all six parts (three sides, three angles) match is overkill: the congruence postulates give shortcuts. SSS (three sides), SAS (two sides and the included angle), ASA (two angles and the included side), and AAS each guarantee congruence. There is deliberately no AAA (similar, not congruent) and no SSA in general (the ambiguous case can produce two different triangles).
CPCTC — Corresponding Parts of Congruent Triangles are Congruent — is the payoff: once triangles are proven congruent, every remaining matching part comes free. This is how geometry proves segment and angle relationships inside complex figures: isolate two triangles, prove them congruent, then read off the needed part. Isosceles triangles get their own theorem: base angles of an isosceles triangle are congruent (and its converse holds too).
The Triangle Inequality says any two sides of a triangle sum to more than the third: sides 3, 4, 8 cannot form a triangle (3 + 4 < 8). The largest angle sits opposite the longest side. The midsegment connecting two sides' midpoints is parallel to the third side and half its length — a favorite source of elegant proofs. And the Pythagorean Theorem, a² + b² = c², with its converse (if a² + b² = c², the triangle is right), remains geometry's most-used result.
Worked example — proving two triangles congruent in a figure:
Given: AB ≅ DE, BC ≅ EF, and ∠B ≅ ∠E (with B between A, C and E between D, F). Prove: ΔABC ≅ ΔDEF.
AB ≅ DE and BC ≅ EF are given (two pairs of congruent sides).
∠B ≅ ∠E is given; ∠B is included between AB and BC, and ∠E between DE and EF (definition of included angle).
Therefore ΔABC ≅ ΔDEF by SAS.
By CPCTC, AC ≅ DF and ∠A ≅ ∠D — bonus results from one congruence proof.
Try it yourself: In ΔABC, AB = AC. Prove ∠B ≅ ∠C (hint: draw the altitude from A and use HL or SSS). Can sides 5, 9, 15 form a triangle? Find the hypotenuse of a right triangle with legs 9 and 12.
Lesson 4: Similar Triangles and Proportions
Similar figures have the same shape but not necessarily the same size: corresponding angles equal, corresponding sides in proportion. Triangles are similar if they satisfy AA (two matching angles), SSS (all sides proportional), or SAS (two sides proportional with the included angle equal). AA is the workhorse: if two angles match, the third must too, since angles sum to 180°.
Similarity solves indirect measurement. A 6-foot person casts a 4-foot shadow while a tree casts a 20-foot shadow at the same moment; the sun's rays make the triangles similar, so height/4 = tree/20... precisely, 6/4 = h/20, giving h = 30 feet. No climbing required. Proportional segments also appear inside triangles: a segment parallel to one side splits the other two sides proportionally.
Be careful to match corresponding sides: in similar triangles ABC ~ DEF, side AB corresponds to DE, not to whichever side looks closest in a rotated diagram. Label carefully, then set up the proportion as a fraction equation and cross-multiply.
Worked example — triangles with sides 3–4–5 and 9–12–x: find x, and check similarity:
Try it yourself: A 5-foot pole casts a 3-foot shadow; a building casts a 45-foot shadow. How tall is the building? Two triangles have sides 4–6–8 and 6–9–13: are they similar? Answers: 5/3 = h/45, h = 75 ft; ratios 6/4 = 1.5, 9/6 = 1.5, 13/8 = 1.625 — not equal, so not similar.
Key vocabulary:Similar — same shape, proportional sides; scale factor — the ratio of corresponding lengths; AA/SSS/SAS similarity — the three tests for triangle similarity.
Unit 2: Polygons, Circles, and Trigonometry
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Lesson 5: Polygons and Quadrilaterals
A polygon is a closed figure of straight segments. Polygons are named by side count: pentagon (5), hexagon (6), octagon (8). A regular polygon has all sides and angles equal; each interior angle of a regular n-gon measures (n − 2) × 180° / n. A regular hexagon's interior angles are (4 × 180)/6 = 120° each — which is why hexagons tile a floor with no gaps.
Quadrilaterals have a family tree. A parallelogram has both pairs of opposite sides parallel; its opposite sides and angles are equal, diagonals bisect each other. A rectangle is a parallelogram with right angles (and equal diagonals); a rhombus is a parallelogram with equal sides (and perpendicular diagonals); a square is both. A trapezoid has exactly one pair of parallel sides; an isosceles trapezoid has non-parallel sides equal and base angles equal. A kite has two pairs of adjacent equal sides.
Area formulas build on the parallelogram: A = base × height for parallelograms (height measured perpendicular to the base, not along the slanted side); triangles take half of that, A = (1/2)bh; trapezoids average the two bases, A = (1/2)(b₁ + b₂)h.
Worked example — area of a trapezoid with bases 10 and 16 and height 7:
A = (1/2)(b₁ + b₂)h = (1/2)(10 + 16) × 7.
(1/2)(26) = 13; 13 × 7 = 91 square units.
Sanity check: between the areas of the 10-by-7 and 16-by-7 rectangles (70 and 112). 91 is plausible.
Try it yourself: (a) Each interior angle of a regular octagon? (b) Area of a parallelogram with base 12 and height 5. (c) A rhombus has diagonals 8 and 10; its area is (1/2)d₁d₂ = ? Answers: (a) (6 × 180)/8 = 135°; (b) 60; (c) 40.
Key vocabulary:Regular polygon — all sides and angles equal; parallelogram — both pairs of opposite sides parallel; trapezoid — exactly one pair of parallel sides.
Lesson 6: Circles
A circle is the set of all points at a fixed distance — the radius — from a center point. Its vocabulary is rich: chord (segment joining two circle points), diameter (chord through the center, twice the radius), tangent (line touching at exactly one point), secant (line cutting through at two points), arc (a curved portion), and central angle (vertex at the center). A tangent is perpendicular to the radius at the point of tangency — the single most-used circle fact in proofs.
Angle chasing in circles runs on a few theorems. An inscribed angle (vertex on the circle) measures half its intercepted arc: an angle subtending a 100° arc is 50°. Corollaries: angles inscribed in the same arc are congruent; an angle inscribed in a semicircle is a right angle (Thales' theorem). The measure of an angle formed by two chords intersecting inside the circle is half the sum of the intercepted arcs; outside, half the difference.
Power and length relationships follow: if two chords intersect inside at P, then (segment 1)(segment 2) of one chord equals that of the other. Arc length is a fraction of the circumference: (central angle/360°) · 2πr. Sector area is the same fraction of the circle's area: (angle/360°) · πr². A 90° sector of radius 6 has area (1/4)π(36) = 9π.
Worked example — finding angles in a circle diagram:
Points A, B, C lie on a circle with center O; arc AC measures 120°. Find the inscribed angle ∠ABC subtending arc AC.
∠ABC is an inscribed angle intercepting arc AC.
Inscribed angle theorem: m∠ABC = ½ · measure of arc AC = ½(120°) = 60°.
Extension: the central angle ∠AOC intercepting the same arc measures the full 120° — central angles equal their arcs.
If D is another point on the major arc AC, then ∠ADC = 60° too (same arc, same angle).
Try it yourself: Two chords intersect inside a circle: one is split into segments 4 and 9, the other into 6 and x. Find x. Find the area of a 60° sector of radius 12, and the length of its arc.
Lesson 7: Right-Triangle Trigonometry
In a right triangle, the ratios of sides are determined entirely by the acute angles — this is trigonometry. For an acute angle θ: sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent (mnemonic: SOH-CAH-TOA). Because similar right triangles share angles, these ratios are the same for every right triangle containing θ; sin(30°) = 1/2 whether the triangle is tiny or huge.
Memorize the special triangles: 30°-60°-90° has sides in ratio 1 : √3 : 2 (short leg : long leg : hypotenuse), and 45°-45°-90° has ratio 1 : 1 : √2. These give exact values — sin(45°) = √2/2, cos(60°) = 1/2 — no calculator needed. The inverse trig functions (arcsin, arccos, arctan) reverse the question: given a ratio, find the angle. If tan(θ) = 5/12, then θ = arctan(5/12) ≈ 22.6°.
Trigonometry solves real measurement problems: heights of buildings (angle of elevation), distances across rivers, navigation bearings. The angle of elevation (looking up) equals the angle of depression (looking down) by alternate interior angles — a surveyor measuring a 40° angle of depression to a boat knows the boat's angle of elevation to her is also 40°. Always sketch, label opposite/adjacent/hypotenuse relative to the known angle, then choose the ratio containing the knowns and the unknown.
Worked example — height of a tree from 30 m away with a 35° angle of elevation:
Sketch: right triangle, adjacent side = 30 m (ground), angle at observer = 35°, opposite = tree height h.
The ratio linking opposite and adjacent is tangent: tan(35°) = h/30.
Solve: h = 30 · tan(35°) ≈ 30 · 0.7002 ≈ 21.0 m.
Add the observer's eye height (say 1.7 m) for the true height: ≈ 22.7 m.
Reasonableness: 35° < 45°, so h < 30 m. Checks out.
Try it yourself: A 10-m ladder leans at 70° to the ground. How high up the wall does it reach? In right ΔABC (∠C = 90°), AC = 7, BC = 24: find sin A, cos A, tan A, and the hypotenuse.
Lesson 8: The Law of Sines and the Law of Cosines
Right-triangle trig (SOHCAHTOA) fails for oblique triangles, so we need two more laws. The Law of Sines states a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius. Use it when you know an angle-side pair plus one more piece (AAS, ASA, or the SSA ambiguous case). The Law of Cosines, c² = a² + b² − 2ab cos C, is the Pythagorean theorem with a correction term; use it when you know two sides and the included angle (SAS) or all three sides (SSS).
The ambiguous case (SSA) is the trap: given two sides and a non-included angle, there may be 0, 1, or 2 triangles. For a = 10, b = 14, A = 30°: sin B = (b sin A)/a = 14(0.5)/10 = 0.7, so B ≈ 44.4° or B ≈ 135.6° (since sin θ = sin(180° − θ)). Both can work if the angles still sum under 180°: with A = 30°, both B values leave a positive third angle, so two triangles exist. Always test the supplementary angle.
These laws power surveying, navigation, and engineering: a surveyor measures a baseline and two angles to locate an inaccessible point; GPS receivers solve triangles between satellites and the ground.
Worked example — triangle with a = 8, b = 10, C = 60°. Find c:
SAS case: use the Law of Cosines: c² = 8² + 10² − 2(8)(10)cos 60°.
cos 60° = 0.5, so c² = 64 + 100 − 160(0.5) = 164 − 80 = 84.
c = √84 ≈ 9.17.
Try it yourself: In a triangle, A = 40°, B = 60°, a = 12. Find b using the Law of Sines. Answers: b = 12 sin 60°/sin 40° ≈ 12(0.866)/0.643 ≈ 16.2.
Key vocabulary:Law of Sines — a/sin A = b/sin B = c/sin C; Law of Cosines — c² = a² + b² − 2ab cos C; ambiguous case — SSA can yield 0, 1, or 2 triangles.
Unit 3: Measurement, Coordinate Geometry, and Transformations
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Lesson 9: Area, Surface Area, and Volume
Two-dimensional area formulas form a family: rectangle lw, parallelogram bh, triangle ½b h, trapezoid ½(b₁+b₂)h, circle πr². Notice the pattern — each is a variation on base-times-height, halved when the shape is “half” of another (triangle from parallelogram). Composite figures yield to decomposition: split into known shapes, compute each, add (or subtract holes).
In three dimensions, surface area is the total area of all faces — the wrapping-paper question — while volume measures capacity. Prisms and cylinders: V = (base area) · height; surface area = 2(base area) + (perimeter of base)(height). Pyramids and cones hold exactly one-third of their prism/cylinder counterparts: V = (1/3)(base area)(height). A sphere: V = (4/3)πr³, SA = 4πr².
The scale factor rules are the lesson's deep idea: scaling all lengths by k multiplies areas by k² and volumes by k³. Double a cube's edge (k=2): surface area quadruples, volume octuples. This is why ants can lift 50 times their weight but elephants can't (strength ~ area, weight ~ volume), and why small mammals lose heat faster (high surface-to-volume ratio). Similar solids' volumes are in the ratio of the cubes of corresponding lengths.
Worked example — a cylindrical water tank, radius 2 m, height 5 m:
Surface area (closed tank): 2πr² + 2πrh = 2π(4) + 2π(2)(5) = 8π + 20π = 28π ≈ 88.0 m².
Open-top version: subtract one base: 28π − 4π = 24π ≈ 75.4 m² of steel.
Scale check: doubling radius and height (k=2) would multiply volume by 8: 160π ≈ 502.7 m³.
Try it yourself: Find the volume of a cone with radius 6 cm and height 10 cm. A sphere has volume 288π: find its radius. Two similar pyramids have heights 4 and 10: what is the ratio of their volumes?
Lesson 10: Coordinate Geometry — Distance, Midpoint, and Slope
Geometry on the coordinate plane turns shapes into algebra. The distance formula, d = √((x₂ − x₁)² + (y₂ − y₁)²), is just the Pythagorean theorem: the horizontal and vertical gaps are the legs, the segment is the hypotenuse. The midpoint formula averages the coordinates: ((x₁ + x₂)/2, (y₁ + y₂)/2). The midpoint of (1, 2) and (7, 8) is (4, 5).
Slopes classify shapes. A quadrilateral with both pairs of opposite sides having equal slopes is a parallelogram; perpendicular adjacent sides (slopes that are negative reciprocals) make it a rectangle. To prove a triangle is a right triangle, check that two sides have slopes whose product is −1. To find a circle's equation from center (h, k) and radius r: (x − h)² + (y − k)² = r² — again the distance formula in disguise.
Coordinate geometry also finds perimeters and areas of plotted shapes, and it locates centers: the centroid (balance point) of a triangle is the average of its three vertices' coordinates.
Worked example — is the triangle with vertices (0, 0), (4, 0), (0, 3) a right triangle? Its area?
Sides: (0,0)–(4,0) is horizontal, length 4. (0,0)–(0,3) is vertical, length 3. Horizontal meets vertical at a right angle, so yes, right triangle.
Hypotenuse: √(4² + 3²) = 5 (3–4–5 triangle).
Area = (1/2)(4)(3) = 6 square units.
Try it yourself: Find the distance and midpoint between (−2, 3) and (4, −5). Answers: d = √(36 + 64) = 10; midpoint = (1, −1).
Key vocabulary:Distance formula — √((Δx)² + (Δy)²); midpoint formula — average of coordinates; centroid — a triangle's balance point, average of vertices.
Lesson 11: Transformations — Reflections, Rotations, and Translations
A transformation moves or resizes a figure. The rigid motions preserve size and shape: translations slide every point by (x + a, y + b); reflections flip across a line (across the x-axis: (x, y) → (x, −y); across the y-axis: (x, y) → (−x, y); across y = x: (x, y) → (y, x)); rotations turn around a point (90° counterclockwise about the origin: (x, y) → (−y, x); 180°: (x, y) → (−x, −y)). A dilation scales by factor k from a center: (x, y) → (kx, ky) from the origin; it preserves shape but changes size.
Congruence and similarity are defined by transformations: two figures are congruent if one can be carried onto the other by rigid motions; similar if rigid motions plus a dilation suffice. This is why SSS, SAS, and ASA prove congruence — they pin down a triangle so completely that a rigid motion must exist carrying one onto the other.
Compositions of transformations have surprising structure: two reflections across parallel lines equal a translation; two across intersecting lines equal a rotation. These facts drive computer graphics, robotics, and animation, where every object on screen is positioned by transformation matrices.
Worked example — triangle A(1, 2), B(4, 2), C(1, 5) reflected across the y-axis, then rotated 180° about the origin:
The final triangle is congruent to the original — rigid motions preserve all lengths.
Try it yourself: Rotate the point (3, 1) 90° counterclockwise about the origin. Reflect (2, −5) across the x-axis. Answers: (−1, 3); (2, 5).
Key vocabulary:Translation / reflection / rotation / dilation — the four basic transformations; rigid motion — preserves distance and angle; congruent — related by rigid motions.
Lesson 12: Dilations and Similarity Transformations
A dilation enlarges or shrinks a figure by a scale factor k, centered at a point: every point's distance from the center is multiplied by k. If k > 1, the figure grows; if 0 < k < 1, it shrinks; if k is negative, the image is also rotated 180° about the center. Dilations preserve shape, angle measures, and parallelism, but change lengths by the factor k and areas by k². A dilation with center at the origin maps (x, y) to (kx, ky).
Similarity transformations are combinations of dilations with rigid motions (translations, reflections, rotations). Two figures are similar exactly when one can be mapped onto the other by a similarity transformation. This is why the side ratios in similar triangles are equal: the dilation sets the ratio, and the rigid motion lines the triangles up. Scale factors also explain maps: a 1:50,000 map means 1 cm on the map is 50,000 cm (0.5 km) in reality.
Worked example — Dilate triangle with vertices (2, 1), (4, 1), (2, 5) by scale factor 3 about the origin:
Multiply each coordinate by 3: (6, 3), (12, 3), (6, 15).
Side lengths triple; area multiplies by 9.
Check: original area = (1/2)(2)(4) = 4; image area = (1/2)(6)(12) = 36 = 9 × 4.
Worked example — A photo 4 in by 6 in is enlarged so the width becomes 10 in. What is the scale factor and the new height?:
Scale factor: 10/4 = 2.5.
New height: 6 × 2.5 = 15 in.
Area scale: 2.5² = 6.25 times the original area.
Real-world application: Photocopiers, maps, blueprints, and phone screens all use dilations. A camera's zoom is a dilation of the image; architects build scale models at a fixed scale factor from the real building.
Try it yourself: Dilate (3, −2) by scale factor 4 about the origin. A model car at 1:24 scale is 8 cm long. How long is the real car? Answers: (12, −8); 8 × 24 = 192 cm = 1.92 m.
Key vocabulary:Dilation — a scaling transformation; scale factor — the multiplier k; similarity transformation — a dilation combined with rigid motions.
Unit 4: Three Dimensions, Constructions, and Analytic Geometry
Three-dimensional solids have volume (space inside) and surface area (skin outside). For a prism or cylinder, V = Bh (base area times height). For a pyramid or cone, V = (1/3)Bh — exactly one-third of the matching prism, a fact you can verify by filling a pyramid with sand three times into its prism. For a sphere, V = (4/3)πr³ and surface area = 4πr². Cavalieri's principle says that if two solids have the same height and the same cross-sectional area at every level, they have the same volume — which is why a slanted prism has the same volume as a straight one.
Surface area of a cone uses the slant height l: lateral area = πrl, plus base πr². For a pyramid, the lateral faces are triangles. When a solid is scaled by factor k, lengths scale by k, areas by k², and volumes by k³ — the reason a model twice as tall holds eight times as much.
Worked example — Volume of a cone with radius 3 cm and height 9 cm; surface area with slant height l = √90 ≈ 9.49 cm:
V = (1/3)πr²h = (1/3)π(9)(9) = 27π ≈ 84.8 cm³.
Lateral area = πrl = π(3)(9.49) ≈ 89.4 cm². Base = 9π ≈ 28.3 cm².
Total surface area ≈ 117.7 cm².
Worked example — A sphere has radius 6 cm. Find its volume and surface area.:
V = (4/3)π(216) = 288π ≈ 904.8 cm³.
SA = 4π(36) = 144π ≈ 452.4 cm².
Note: doubling the radius would multiply volume by 8.
Real-world application: Engineers size water tanks (cylinders and spheres), architects design domes, and manufacturers minimize packaging surface area for a fixed volume — a sphere is the most efficient shape, which is why bubbles and planets are round.
Try it yourself: A rectangular prism is 2 by 3 by 10 cm. (a) Find its volume. (b) A pyramid with the same base and height has what volume? Answers: (a) 60 cm³; (b) 20 cm³ (one-third).
Key vocabulary:Cavalieri's principle — equal heights and equal cross-sections imply equal volumes; slant height — the distance from a cone's apex to the base edge along the side.
Lesson 14: Geometric Constructions with Compass and Straightedge
A construction is a drawing made using only a compass and an unmarked straightedge — no measuring, no protractors. The compass transfers distances (it can draw a circle of a given radius), and the straightedge draws lines through points. From just these tools come all the basic constructions: copying a segment, copying an angle, bisecting a segment, bisecting an angle, drawing a perpendicular through a point, and drawing a parallel through a point.
Constructions are proofs in action: the perpendicular bisector construction works because its points are equidistant from the segment's endpoints (the compass guarantees it), and the angle bisector works by SSS congruence of the two triangles formed. Constructing a regular hexagon is easy (six radii around a circle); constructing a regular heptagon with these tools alone is impossible — a fact proved in the 19th century.
Worked example — Construct the perpendicular bisector of segment AB:
Set the compass wider than half of AB. Draw arcs centered at A and at B; they intersect at points P and Q.
Draw line PQ. It crosses AB at its midpoint M, at a right angle.
Why it works: P and Q are equidistant from A and B, so both lie on the perpendicular bisector; two points determine the line.
Worked example — Construct an angle bisector of angle C:
Draw an arc centered at C cutting both sides at points D and E.
With the same compass width, draw arcs centered at D and E; they meet at F.
Draw ray CF — it bisects the angle (triangles CDF and CEF are congruent by SSS).
Real-world application: Stonemasons, carpenters, and engineers still use compass-and-straightedge thinking to lay out arches, bisect corners, and find centers of circles without measuring tools. CAD software's snap and mirror commands are digital versions of the same constructions.
Try it yourself: Describe how to copy an angle using a compass and straightedge. Answer: Draw an arc across the original angle meeting its sides at two points; draw a ray with the same arc radius from a point on it; transfer the chord distance between the two points with the compass; the ray through the new point copies the angle.
Key vocabulary:Construction — a figure drawn with only compass and straightedge; perpendicular bisector — the line of points equidistant from a segment's endpoints; bisect — to divide into two equal parts.
Lesson 15: Coordinate Geometry: Equations of Circles and Lines
Coordinate geometry turns geometry into algebra. A circle with center (h, k) and radius r is the set of points at distance r from the center, so by the distance formula its equation is (x − h)² + (y − k)² = r². To find a circle's center from a general equation, complete the square in x and y. A line through (x1, y1) with slope m is y − y1 = m(x − x1); parallel lines share slopes, perpendicular lines have slopes that are negative reciprocals.
Coordinate proofs use these tools to prove geometric facts with algebra. To prove the diagonals of a rectangle bisect each other, place the rectangle at (0, 0), (a, 0), (a, b), (0, b), and show both diagonals have the same midpoint, (a/2, b/2). Placing the figure with a vertex at the origin and sides on the axes makes the algebra simplest — a strategic choice, not a coincidence.
Worked example — Find the center and radius of x² + y² + 8x − 6y + 16 = 0:
Worked example — Write the equation of the line perpendicular to y = 2x + 1 through (4, −3):
Perpendicular slope: −1/2 (negative reciprocal of 2).
Point-slope: y + 3 = −(1/2)(x − 4).
Simplify: y = −(1/2)x − 1.
Real-world application: GPS navigation, video-game collision detection, and architectural drafting all live in coordinate geometry: circles are radar ranges, lines are paths, and midpoints are meeting points.
Try it yourself: Find the equation of the circle with center (2, −5) and radius 7. Is the point (3, 1) on it? Answers: (x − 2)² + (y + 5)² = 49; (3−2)² + (1+5)² = 1 + 36 = 37 ≠ 49, so no.
Key vocabulary:Coordinate proof — proving a geometric fact with algebra; negative reciprocal — the slope −1/m perpendicular to m.
Lesson 16: Geometric Probability
Geometric probability replaces counting outcomes with measuring lengths, areas, or volumes: P = (measure of the favorable region) / (measure of the whole region). If a dart lands at random on a 10-by-10 board and the bullseye is a circle of radius 2, the probability of hitting it is the circle's area divided by the board's area: 4π/100 ≈ 12.6%. The key assumption is that every point is equally likely — the dart has no favorite spot.
On a line segment, probability is a ratio of lengths: if a bus arrives at a random time in a 30-minute window and you wait 10 minutes, the probability it arrives while you wait is 10/30 = 1/3. Geometric probability also handles “meeting” problems: two friends arriving at random times in an hour, each waiting 15 minutes, meet with probability 1 − (45/60)² = 7/16, from the area of a square minus two triangles.
Worked example — A dart lands randomly on a 20 cm by 20 cm board. What is P(landing in a 4 cm by 4 cm square in the corner)?:
Favorable area: 16 cm². Total area: 400 cm².
P = 16/400 = 0.04 = 4%.
The ratio of areas is the probability.
Worked example — A point is chosen at random on a segment 12 cm long. What is P(it lies within 3 cm of an endpoint)?:
Favorable length: 3 cm at each end = 6 cm total.
P = 6/12 = 1/2.
On a line, probability is a ratio of lengths.
Real-world application: Archers' scoring rings, rainfall over a region, quality-control sampling, and collision risk for satellites are all geometric probability: favorable area or volume over total area or volume.
Try it yourself: A circular target has radius 10 cm; the bullseye has radius 2 cm. A dart hitting the target at random: P(bullseye)? Answer: Areas: 4π/100π = 1/25 = 4%.
Key vocabulary:Geometric probability — probability as a ratio of measures (length, area, volume); uniform — equally likely everywhere in the region.
Unit 5: Advanced Proof, Symmetry, and Capstone
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Lesson 17: Logic and Advanced Proofs
Beyond basic congruence proofs, two powerful techniques solve harder problems. Indirect proof (proof by contradiction): assume the opposite of what you want to prove, and derive a contradiction — so the assumption must be false. To prove a triangle cannot have two right angles, assume it does, and the angle sum exceeds 180°, contradicting the triangle angle-sum theorem.
CPCTC (Corresponding Parts of Congruent Triangles are Congruent) extends a congruence proof: once triangles are congruent, every matching side and angle is equal. Overlapping triangles — two triangles sharing a side or angle — are a classic CPCTC setup: prove the overlap triangles congruent, then conclude the target parts are equal. Other tools: the triangle inequality (the sum of any two sides exceeds the third), the hinge theorem, and properties of isosceles triangles (base angles equal).
Worked example — Prove: If two angles of a triangle are equal, the sides opposite them are equal (converse of the base angles theorem):
Let triangle ABC with ∠A = ∠B. Draw the angle bisector from C to side AB at D.
Triangles ACD and BCD share CD, have ∠A = ∠B, and ∠ACD = ∠BCD by construction — congruent by AAS.
By CPCTC, AD = BD is not the claim; instead, AC = BC by CPCTC. The triangle is isosceles.
Worked example — Can a triangle have sides 4, 7, and 12?:
Triangle inequality: 4 + 7 = 11 < 12, so the two short sides cannot meet.
No such triangle exists. The sum of any two sides must exceed the third.
Real-world application: Lawyers use proof by contradiction; engineers prove a structure cannot fail in a certain way by assuming it does and finding the contradiction; software verification proves programs correct with the same logic.
Try it yourself: Use an indirect proof: prove that a triangle cannot have two obtuse angles. Answer: Assume it has two obtuse angles (> 90° each). Their sum exceeds 180°, but all three angles sum to exactly 180° and the third is positive — contradiction. So at most one obtuse angle.
Key vocabulary:Indirect proof — assuming the negation and deriving a contradiction; CPCTC — corresponding parts of congruent triangles are congruent; triangle inequality — the sum of two sides exceeds the third.
Lesson 18: Symmetry and Tessellations
Symmetry is a transformation that leaves a figure unchanged. Line (reflection) symmetry means the figure folds onto itself along a line; a square has 4 lines of symmetry, a regular hexagon 6, a circle infinitely many. Rotational symmetry means the figure maps onto itself after a rotation of less than 360°; a square has 90° rotational symmetry (order 4). A figure can have one, both, or neither.
A tessellation (tiling) covers a plane with shapes, no gaps or overlaps. The three regular tessellations use equilateral triangles, squares, and regular hexagons — the only regular polygons whose interior angle divides 360° (60°, 90°, 120°). Semi-regular tessellations mix two or more regular polygons, and Escher's art shows tessellations with curved, interlocking shapes. The angle condition is the key: at every vertex, the angles meeting must sum to exactly 360°.
Worked example — How many lines of symmetry does a rectangle (not a square) have? What is its rotational symmetry?:
Lines of symmetry: 2 (through the midpoints of opposite sides). The diagonals are NOT lines of symmetry for a non-square rectangle.
Rotational symmetry: 180° (order 2).
A square adds the two diagonals for 4 lines and 90° rotation.
Worked example — Can regular pentagons tessellate the plane?:
Interior angle of a regular pentagon: (5 − 2)(180)/5 = 108°.
360/108 = 3.33 — not a whole number, so pentagons cannot meet at a vertex without a gap or overlap.
Regular pentagons do not tessellate.
Real-world application: Tile floors, honeycombs (hexagons), brick walls, and quilt patterns are tessellations. Architects use symmetry to design balanced buildings, and logos use symmetry to look stable and professional.
Try it yourself: How many lines of symmetry does an equilateral triangle have? Does a regular octagon tessellate alone? Answers: 3 lines, 120° rotational symmetry (order 3); octagon interior angle 135°, 360/135 is not whole, so no (but octagons + squares tessellate).
Key vocabulary:Line symmetry — reflection maps the figure onto itself; rotational symmetry — rotation maps the figure onto itself; tessellation — a gapless, overlap-free tiling of the plane.
Lesson 19: Geometry in the Real World: Design, Architecture, and Maps
Geometry is the mathematics of the built world. Scale drawings and blueprints use a scale factor: a 1:48 blueprint means 1 unit on paper is 48 units in the building, so a 3-inch wall on paper is 144 inches (12 feet) in reality. Scale factor also governs area (k²) and volume (k³): a model at 1:10 scale has 1/100 the surface area and 1/1000 the volume of the real object.
Triangulation locates points from angles and a known baseline — the principle behind surveying and GPS. In a building, load paths follow triangles because triangles are rigid: a rectangle can collapse into a parallelogram, but a triangle cannot change shape without changing a side length. That is why bridges, roofs, and cranes are full of triangular trusses. Perspective drawing uses vanishing points where parallel lines appear to meet, turning three-dimensional scenes into two-dimensional images.
Worked example — A blueprint uses a scale of 1 inch = 4 feet. A room measures 3.5 in by 2.25 in on the blueprint. What are its real dimensions and area?:
Real length: 3.5 × 4 = 14 ft. Real width: 2.25 × 4 = 9 ft.
Real area: 14 × 9 = 126 ft².
Check by area scale: drawing area 7.875 in² × 16 (ft² per in²) = 126 ft².
Worked example — Two surveyors 100 m apart measure angles of 60° and 50° to a tree. How far is the tree from the first surveyor? (Use the Law of Sines.):
The triangle's third angle: 180 − 60 − 50 = 70°.
By the Law of Sines: d / sin 50° = 100 / sin 70°, so d = 100 · sin 50° / sin 70° ≈ 100 · 0.766/0.940 ≈ 81.5 m.
Triangulation gives a distance nobody walked.
Real-world application: Architects, surveyors, game designers, and civil engineers use scale, triangulation, and triangular rigidity daily. The same geometry that proves a theorem designs a bridge.
Try it yourself: A map scale is 1 cm = 25 km. Two cities are 8.4 cm apart on the map. How far apart are they really? Answer: 8.4 × 25 = 210 km.
Key vocabulary:Scale drawing — a drawing with a fixed ratio to the real object; triangulation — locating a point from angles and a baseline; truss — a rigid framework of triangles.
Lesson 20: Geometry Capstone: Proof Portfolio and Review
The capstone gathers the course into a proof portfolio: a collection of your best proofs, constructions, and applications, organized by theme — congruence, similarity, circles, area and volume, and coordinate geometry. A strong portfolio entry states the given and the goal, shows a labeled diagram, gives a two-column proof or a clear paragraph proof, and ends with a one-sentence reflection: what made this proof work?
Review problem 1 — Similarity in a right triangle. The altitude to the hypotenuse of a right triangle creates three similar triangles; the geometric mean relations follow: the altitude h satisfies h² = pq, where p and q are the segments of the hypotenuse. Review problem 2 — Coordinate proof. Show the diagonals of a parallelogram bisect each other by placing it at (0, 0), (a, 0), (a + b, c), (b, c) and computing midpoints. The portfolio's final entry is a real-world project: measure a tall object using shadows and similar triangles, or design a tessellation, and present the geometry behind it.
Worked example — The altitude to the hypotenuse is 6 and divides the hypotenuse into segments 4 and x. Find x.:
Geometric mean relation: h² = pq, so 36 = 4x.
x = 9. The hypotenuse is 13.
Check: the legs are √(4·13) = √52 and √(9·13) = √117; 52 + 117 = 169 = 13².
Worked example — Find the area of a regular hexagon with side 6.:
A regular hexagon splits into 6 equilateral triangles of side 6.
Each triangle's area: (√3/4)(36) = 9√3.
Total: 54√3 ≈ 93.5.
Real-world application: The capstone habits — diagram, given/goal, logical chain, reflection — are exactly the habits of engineers writing a design justification and scientists writing a proof of a result. Geometry ends with the ability to argue precisely about shape and space.
Try it yourself: A right triangle has legs 9 and 12. Find the hypotenuse and the altitude to the hypotenuse. Answers: Hypotenuse 15; altitude h satisfies (1/2)(9)(12) = (1/2)(15)h, so h = 108/15 = 7.2.
Key vocabulary:Proof portfolio — a curated collection of proofs and applications; geometric mean — for the altitude relation, h = √(pq).
Key vocabulary
Postulate
A statement accepted without proof, serving as a starting point for reasoning.
Theorem
A statement proven true from postulates, definitions, and other theorems.
Contrapositive
The logically equivalent form 'if not q, then not p' of 'if p, then q'.
Congruent
Identical in shape and size; one figure maps exactly onto the other.
CPCTC
Corresponding Parts of Congruent Triangles are Congruent — the payoff of a congruence proof.
Hypotenuse
The side opposite the right angle; the longest side of a right triangle.
Tangent (circle)
A line touching a circle at exactly one point, perpendicular to the radius there.
Inscribed angle
An angle with vertex on the circle; measures half its intercepted arc.
Sine
In a right triangle, opposite/hypotenuse for a given acute angle.
Cosine
In a right triangle, adjacent/hypotenuse for a given acute angle.
Tangent (trig)
In a right triangle, opposite/adjacent for a given acute angle.
Volume
The three-dimensional space a solid occupies, in cubic units.
Surface area
The total area of all faces of a solid, in square units.
Scale factor
The multiplier relating corresponding lengths of similar figures; areas scale by its square, volumes by its cube.
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The four triangle families, classified by sides and angles.
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