Around 300 BCE, Euclid of Alexandria set out to do something unusual. Instead of collecting useful facts about shapes, which the Egyptians and Babylonians had been doing for over a thousand years, he wanted every fact to follow from a handful of statements so obvious that nobody could reasonably reject them. If you accept the starting rules, you must accept everything built from them.

That is exactly how a software engineer thinks about a type system or a set of database constraints: fix a small set of axioms, and every guarantee downstream is earned rather than assumed. Geometry was the first place humans did this systematically.

Three words are undefined: we describe them, but we don’t define them in terms of anything simpler. Every other term is defined from them.

Euclid’s five postulates, in modern language:

The first four look like rules for using a ruler and compass. The fifth is different in character, and it is the one that powers most of this chapter: the angle sum of a triangle, similarity, and the Pythagorean theorem all depend on it.

Two roads cross. One makes a 63° angle with the other. What are the other three angles at the junction? Now a third road runs parallel to the first. What angles does it make with the second road? You should be able to answer both without a protractor, and the reasons are worth knowing because they are the first links in every geometric proof.

Vertical angles are equal. When two lines cross, angles 1 and 2 form a straight line, so \(\angle 1 + \angle 2 = 180^\circ\). Angles 2 and 3 also form a straight line, so \(\angle 2 + \angle 3 = 180^\circ\). Both sums equal 180°, so \(\angle 1 = 180^\circ - \angle 2 = \angle 3\). That is the whole proof; we’ll write it formally in Section 9.9.

Parallel lines cut by a transversal. A transversal is a line crossing two or more other lines. When it crosses two parallel lines it creates eight angles, and the parallel postulate forces the two crossings to look identical: the transversal meets \(\ell_2\) at exactly the same tilt as \(\ell_1\). So the four angles at the top crossing are copies of the four at the bottom.

The pairs have names, which US tests use heavily:

The table is really one fact: every acute angle in the figure is equal, every obtuse angle is equal, and an acute plus an obtuse makes 180°. (If the transversal is perpendicular, all eight are right angles.) The UK mnemonic is F-angles (corresponding), Z-angles (alternate) and C- or U-angles (co-interior), after the letter shapes they trace.

The facts also run backwards, and that is what makes them useful: if a transversal makes equal corresponding angles with two lines, those lines are parallel. This converse is how you prove lines parallel.

Two parallel lines are cut by a transversal; angle 1 in the figure measures \(63^\circ\). Find all the others.

Check: \(\angle 4 + \angle 5 = 117^\circ + 63^\circ = 180^\circ\), as same-side interior angles must.

A typical algebraic version: same-side interior angles measure \((3x + 15)^\circ\) and \((2x - 5)^\circ\). Since they are supplementary,

so the angles are \(3(34) + 15 = 117^\circ\) and \(2(34) - 5 = 63^\circ\).

Navigation (bearings). A bearing is an angle measured clockwise from north. North lines at two different places are parallel, so the path between them is a transversal. If a boat sails from \(A\) to \(B\) on a bearing of \(063^\circ\), the angle at \(B\) between north and the way back is the co-interior partner of \(63^\circ\), namely \(117^\circ\) measured anticlockwise, so the return bearing is \(360^\circ - 117^\circ = 243^\circ\). That’s the rule every navigator and orienteer uses: back bearing = bearing \(\pm 180^\circ\).

DIY. A picture frame’s corners are \(90^\circ\), so each of the two pieces meeting at a corner is cut at \(45^\circ\) (the cut bisects the corner). A hexagonal mirror frame has interior angles of \(120^\circ\), so each cut is at \(60^\circ\) to the edge of the strip, which a mitre saw (whose dial reads 0° for a square cut) calls \(30^\circ\).

Physics. Light reflects off a mirror at the same angle it arrives. Put two mirrors at a right angle and a ray that hits both comes back exactly parallel to the way it came, whatever angle it arrived at: the two reflections turn it through \(2 \times 90^\circ = 180^\circ\). Bicycle reflectors, road "cat’s eyes" and the reflector arrays the Apollo astronauts left on the Moon (used to measure the Earth–Moon distance by laser to within centimetres) are three-mirror versions of this corner.

Computing. In a 2D game, a ball bouncing off a vertical wall keeps its vertical velocity and flips the sign of its horizontal velocity, vx = -vx. Equal angles in and out is the geometric fact; one sign flip is its implementation. Ray tracers do the same thing for every reflective surface, using vectors (Chapter 26).

Why do the angles of every triangle add to 180°? You’ve known the fact since childhood, but it’s not obvious: a long thin triangle and a fat equilateral one look nothing alike. And a related engineering question: why are bridges and roof trusses built from triangles rather than rectangles? Both answers come from the same place.

Theorem. The interior angles of any triangle add to \(180^\circ\).

Proof. Take \(\triangle ABC\) with angles \(\alpha\) at \(A\), \(\beta\) at \(B\), \(\gamma\) at \(C\). Through \(C\), draw the line parallel to \(\overline{AB}\) (the parallel postulate says there is exactly one).

Side \(\overline{AC}\) is a transversal of the two parallel lines, so the angle between the new line and \(\overline{CA}\) equals \(\alpha\) (alternate interior angles). Likewise \(\overline{BC}\) is a transversal, so the angle on the right at \(C\) equals \(\beta\). Now the three angles at \(C\), namely \(\alpha\), \(\gamma\), \(\beta\), sit side by side along a straight line, so

Notice what the proof used: the parallel postulate and the alternate-angle fact. On a sphere, where the parallel postulate fails, so does the theorem.

Corollary (exterior angle theorem). An exterior angle of a triangle equals the sum of the two interior angles not adjacent to it. If the exterior angle at \(C\) is \(\delta\), then \(\delta = 180^\circ - \gamma = \alpha + \beta\).

Triangle inequality. Any side of a triangle is shorter than the sum of the other two: \(AB < AC + CB\). "The straight path is the shortest." Lengths 2, 3 and 6 can’t form a triangle because \(2 + 3 < 6\). This inequality reappears as a defining property of every distance function in mathematics and computing.

Two figures are congruent (\(\cong\)) if one can be moved onto the other exactly by sliding, rotating and possibly flipping it. Same shape, same size. For triangles, all six parts (three sides, three angles) match.

The practical question: how little do you need to check to know two triangles are congruent? Think of it as "how much information pins down a triangle uniquely". If you’re told three side lengths, you can build only one triangle (up to moving it around): try it with three sticks. That is why a triangle frame is rigid while a rectangular frame can be pushed into a parallelogram. Four sides do not determine a quadrilateral; three sides do determine a triangle. Hence roof trusses.

Try it: See both SSA triangles.

In \(\triangle PQR\) and \(\triangle XYZ\): \(PQ = XY = 7\), \(\angle Q = \angle Y = 50^\circ\), \(QR = YZ = 9\). Are they congruent, and if so what is \(\angle R\) given \(\angle Z = 72^\circ\)?

Home. A flat-pack bookcase that sways sideways is a rectangle, which can shear into a parallelogram without changing any side length. The thin board nailed across the back turns it into two triangles, and SSS says a triangle with fixed sides can’t change shape. The same reason puts diagonal braces in garden gates (the brace runs from the bottom of the hinge side up to the top of the latch side, so it is pushed, not pulled) and triangles in every bicycle frame.

Engineering and surveying. Bridges and roof trusses (Warren and Pratt trusses are rows of triangles) and electricity pylons are triangulated for rigidity. Before satellites, national maps such as the UK Ordnance Survey were built from a network of triangles between hilltop "trig pillars": measure one baseline very accurately, measure angles, and the angle facts and similarity fix every other point.

Route planning and search. The triangle inequality gives you bounds for free. If a road trip is 30 km from \(A\) to \(B\) and 40 km from \(B\) to \(C\), then \(A\) and \(C\) are at least \(40 - 30 = 10\) km and at most \(70\) km apart in a straight line. Route planners such as A* use the straight-line distance as a lower bound on the road distance, which is what lets them skip hopeless routes safely. Nearest-neighbour indexes for search use the same trick: if your query is 10 units from a reference point \(p\), and an item is 3 units from \(p\), that item is at least \(10 - 3 = 7\) units from your query, so if you already have a match at distance 5 you never need to compute that item’s distance at all.

You want the height of a tree. You can’t climb it, but it’s a sunny day. You are 1.8 m tall and cast a 2.4 m shadow; the tree casts a 16 m shadow. How tall is the tree? Thales of Miletus supposedly measured the Great Pyramid this way around 600 BCE.

Two figures are similar if one is a scaled copy of the other: same shape, possibly different size, possibly moved, rotated or flipped. Every length is multiplied by the same number \(k\), the scale factor, and every angle is unchanged. A photo and its thumbnail are similar; a photo stretched to fill a different aspect ratio is not.

In the tree problem, the sun’s rays are parallel, so the triangle formed by you, your shadow and the ray to the tip of your shadow has the same angles as the corresponding triangle for the tree. Same angles means same shape, so the ratios of sides are equal.

AA is the workhorse: angles are usually easier to get than lengths, and parallel lines hand them to you for free. For instance, a line drawn across a triangle parallel to one side cuts off a smaller triangle similar to the original (the corresponding angles are equal). That fact is called the side-splitter or basic proportionality theorem.

If lengths scale by \(k\), what happens to area and volume?

A square of side \(s\) has area \(s^2\). Scale it by \(k\) and its area is \((ks)^2 = k^2 s^2\). Any shape can be approximated by tiny squares, each of which scales by \(k^2\), so:

The 3-4-5 triangle has area \(\tfrac12 \cdot 3 \cdot 4 = 6\); the 6-8-10 triangle has area \(\tfrac12 \cdot 6 \cdot 8 = 24 = 2^2 \times 6\).

Both triangles have a right angle (person and tree are vertical) and the same angle of elevation of the sun. By AA they are similar. Matching "height ↔ height" and "shadow ↔ shadow":

Sanity check: the scale factor is \(k = 16 / 2.4 = 6.\overline{6}\), and \(1.8 \times 6.\overline{6} = 12\). The tree’s shadow is much longer than yours, and so is the tree.

Alternative approach. Use the ratio within one triangle instead of between triangles: your height-to-shadow ratio is \(1.8 / 2.4 = 0.75\), and the same ratio holds for the tree, so \(h = 0.75 \times 16 = 12\). This "within" ratio is precisely what trigonometry will name the tangent of the sun’s elevation angle (Chapter 13). Choosing a method: use "between" ratios when a scale factor is given or obvious (maps, models); use "within" ratios when you have a single triangle’s shape and want to apply it to another, which is the trigonometric view.

A 5 m ladder leans against a wall with its foot 1.4 m from the wall. How high up the wall does it reach? A television is sold as "55-inch" (the diagonal) with a 16:9 aspect ratio; how wide is it? A game character is at \((3, 4)\) relative to an enemy; how far away is that? All three questions are the same question.

There are hundreds of proofs. This one is the most convincing at a glance.

Take four copies of the right triangle with legs \(a, b\) and hypotenuse \(c\). Arrange them inside a square of side \(a + b\) in two ways.

The big square and the four triangles are identical in both pictures, so the uncovered areas must match: \(c^2 = a^2 + b^2\). \(\blacksquare\)

The same picture gives an algebraic proof. Count the left-hand big square two ways:

Why the converse is true. Suppose \(\triangle ABC\) has \(a^2 + b^2 = c^2\). Build a genuine right triangle with legs \(a\) and \(b\); by the theorem its hypotenuse is \(\sqrt{a^2 + b^2} = c\). The two triangles now agree in all three sides, so they are congruent by SSS, and the original angle opposite \(c\) is a right angle too.

The ladder. The ladder is the hypotenuse (\(c = 5\)); one leg is \(b = 1.4\); we want the height \(a\). Try it: Solve the ladder triangle.

(This is the 7-24-25 triple scaled by \(0.2\): \(1.4, 4.8, 5\).)

The television. With aspect ratio 16:9, write width \(= 16t\) and height \(= 9t\) for some \(t\). The diagonal is 55:

so the width is \(16t \approx 47.9\) inches and the height \(9t \approx 27.0\) inches.

The converse at work. Is a triangle with sides 9, 40, 41 a right triangle? \(9^2 + 40^2 = 81 + 1600 = 1681 = 41^2\). Yes. And 6, 7, 9? \(36 + 49 = 85 \neq 81 = 9^2\), so it isn’t a right triangle. In fact if \(a^2 + b^2 > c^2\) (with \(c\) the longest side) the triangle is acute, and if \(a^2 + b^2 < c^2\) it is obtuse. The law of cosines in Chapter 13 makes this precise.

Two points \((x_1, y_1)\) and \((x_2, y_2)\) are the ends of the hypotenuse of a right triangle whose legs are the horizontal and vertical separations \(\Delta x = x_2 - x_1\) and \(\Delta y = y_2 - y_1\). (\(\Delta\), Greek capital delta, means "change in".) So the distance is

From \((1, 2)\) to \((4, 6)\): \(\sqrt{3^2 + 4^2} = 5\). Chapter 10 develops this fully as the distance formula.

In 3D, apply the theorem twice. The diagonal of a box measuring \(3 \times 4 \times 12\): the floor diagonal is \(\sqrt{3^2 + 4^2} = 5\), and that floor diagonal together with the height forms another right triangle, so the space diagonal is \(\sqrt{5^2 + 12^2} = 13\). In general \(d = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}\), and the pattern continues to any number of dimensions, which is how machine learning measures the distance between 768-dimensional embedding vectors (Chapter 26).

Two right triangles turn up so often that it pays to know their side ratios by heart. Both come from cutting a regular shape in half.

45-45-90. Cut a unit square along its diagonal. Each half is an isosceles right triangle with legs 1 and hypotenuse \(\sqrt{1^2 + 1^2} = \sqrt2\). Sides are in the ratio \(1 : 1 : \sqrt2\).

30-60-90. Cut an equilateral triangle of side 2 down its line of symmetry. Each half has hypotenuse 2, short leg 1 (half the base), and angles \(30^\circ, 60^\circ, 90^\circ\). The long leg is \(\sqrt{2^2 - 1^2} = \sqrt3\). Sides are in the ratio \(1 : \sqrt3 : 2\), with the short leg opposite the \(30^\circ\) angle.

Example. An equilateral triangle has side 10. Its height is the long leg of a 30-60-90 triangle with hypotenuse 10, short leg 5, so the height is \(5\sqrt3 \approx 8.66\) and the area is \(\tfrac12 \cdot 10 \cdot 5\sqrt3 = 25\sqrt3 \approx 43.30\). In general, an equilateral triangle of side \(s\) has area \(\tfrac{\sqrt3}{4}s^2\).

These ratios are exactly the values of sine, cosine and tangent at \(30^\circ, 45^\circ, 60^\circ\) that Chapter 13 will ask you to know: for example \(\sin 30^\circ = \tfrac12\) is "short leg over hypotenuse".

A surveyor walks around a plot of land and records the corners as coordinates. What is its area? A tiler wants to know which regular shapes can tile a floor with no gaps. Both need the same two tools: angle sums of polygons and a reliable way to compute area.

Pick one vertex of a convex \(n\)-gon and draw every diagonal from it. You cut the polygon into \(n - 2\) triangles (each side except the two touching your vertex becomes the base of one triangle). Each triangle contributes \(180^\circ\), so

For a regular polygon each interior angle is \(\dfrac{(n-2) \times 180^\circ}{n}\): 60° (triangle), 90° (square), 108° (pentagon), 120° (hexagon), 135° (octagon).

The exterior angles are even simpler. Walk around any convex polygon: at each corner you turn through the exterior angle, and after one lap you have turned through exactly one full turn. So the exterior angles of every convex polygon add to \(360^\circ\), regardless of \(n\). For a regular \(n\)-gon each exterior angle is \(360^\circ/n\), which is the quickest way to find \(n\): a regular polygon with interior angles of \(156^\circ\) has exterior angles of \(24^\circ\), so \(n = 360/24 = 15\).

Every basic area formula comes from the rectangle by cutting and rearranging, which is worth seeing because it means you never need to memorise them blindly.

Rectangle: \(A = bh\) (base \(b\), height \(h\)). That’s the definition of area in square units: \(b\) columns of \(h\) unit squares.

Parallelogram: cut the triangle off one end and slide it to the other end. You get a rectangle with the same base and the same perpendicular height: \(A = bh\).

Triangle: two copies of any triangle fit together into a parallelogram with the same base and height. So

where \(h\) is the perpendicular height measured to the line containing the base (for an obtuse triangle that may fall outside the triangle).

Trapezoid with parallel sides \(a\) and \(b\), height \(h\): two copies, one rotated, form a parallelogram of base \(a + b\), so \(A = \tfrac12 (a + b) h\). ("Average of the parallel sides, times the height.")

The surveyor’s problem: you have the corners \((x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)\) of a simple polygon, listed in order around the boundary. There’s no obvious base or height. The shoelace formula (or surveyor’s formula) handles any simple polygon, convex or not:

Read the notation piece by piece: \(\sum_{i=1}^{n}\) (capital sigma) means "add up the following expression for \(i = 1, 2, \dots, n\)" (it’s a for loop, and Chapter 14 covers it in detail); \(|\cdot|\) is absolute value; and the condition at the end says "wrap around" so the last vertex pairs with the first.

Why it works (sketch). Each term \(\tfrac12 (x_i y_{i+1} - x_{i+1} y_i)\) is the signed area of the triangle formed by the origin and the edge from vertex \(i\) to vertex \(i+1\): positive if the edge goes anticlockwise around the origin, negative if clockwise. Adding these triangles for every edge, the parts outside the polygon cancel (they are swept once each way) and only the polygon’s area remains. The sign of the total tells you the vertex order: positive for anticlockwise, negative for clockwise. (In Chapter 27 each term turns out to be a \(2 \times 2\) determinant.) The name comes from the criss-cross pattern of multiplications when you write the coordinates in two columns, like lacing a shoe.

Worked example. Vertices \(P_1(1,1)\), \(P_2(5,2)\), \(P_3(4,5)\), \(P_4(2,4)\). Try it: Shoelace area calculator.

The sum is positive, so the vertices were listed anticlockwise.

Alternative approach: box and subtract. Draw the bounding rectangle from \(x = 1\) to \(5\) and \(y = 1\) to \(5\): area 16. Subtract the pieces of the box outside the polygon: below edge \(P_1P_2\), a right triangle \(\tfrac12 \cdot 4 \cdot 1 = 2\); right of edge \(P_2P_3\), \(\tfrac12 \cdot 1 \cdot 3 = 1.5\); above edge \(P_3P_4\), \(\tfrac12 \cdot 2 \cdot 1 = 1\); left of edge \(P_4P_1\), \(\tfrac12 \cdot 1 \cdot 3 = 1.5\); and one piece people forget, the \(1 \times 1\) square in the top-left corner between \(P_4(2,4)\) and the box corner \((1,5)\), which belongs to neither neighbouring triangle. Total removed: \(2 + 1.5 + 1 + 1.5 + 1 = 7\), so \(A = 16 - 7 = 9\). The box method works for simple shapes on grid paper, but that forgotten corner square shows how easy it is to get wrong; the shoelace formula is mechanical and never needs a picture.

Choosing a method. For triangles and standard quadrilaterals with known base and height, use the formulas. For any polygon given by coordinates, use the shoelace formula; it’s exact, handles concave shapes, and is a five-line program. For lattice polygons (all vertices on integer points) there’s also Pick’s theorem: \(A = I + B/2 - 1\), with \(I\) interior lattice points and \(B\) boundary lattice points. Here \(I = 8\) and \(B = 4\) (the edges pass through no other lattice points), so \(A = 8 + 2 - 1 = 9\). Consistent.

Is one 18-inch pizza more food than two 12-inch pizzas? (Most people guess no.) How long is the crust on a 45° slice? And a puzzle that has a lovely answer: if you stand anywhere on the edge of a circular stage and look at two fixed points on the edge, does the angle you see depend on where you stand?

From the definition of \(\pi\):

For the area, cut the circle into many thin sectors and rearrange them alternately point-up and point-down.

The result is almost a parallelogram. Its top and bottom edges are each half the circumference, \(\pi r\), and its height is almost exactly \(r\). As the slices get thinner, the scalloped edges flatten and the shape becomes a true rectangle \(\pi r\) by \(r\). So

"As the slices get thinner, the approximation becomes exact" is the idea of a limit (Chapter 16), and cutting a region into thin pieces and adding them up is exactly what an integral does (Chapter 19). You’ve just done calculus informally, as Archimedes did.

A sector or arc with central angle \(\theta\) (in degrees) is the fraction \(\theta / 360\) of the whole:

The pizza. Diameters 18 and 12 mean radii 9 and 6.

The single large pizza wins by about 12%. Area scales as the square of the diameter (scaling law again), and intuition consistently underestimates squares.

The slice. A 40 cm pizza (\(r = 20\)) cut into eight 45° slices.

Two facts about angles in circles have short, satisfying proofs and many uses.

Tangent–radius theorem. A tangent is perpendicular to the radius at the point of tangency. (Intuition: the radius is the shortest path from the centre to the tangent line, since every other point of the line is outside the circle; and the shortest path from a point to a line is the perpendicular one.)

Inscribed angle theorem. An inscribed angle is half the central angle standing on the same arc.

Proof (the case where the centre lies inside the inscribed angle). Draw the diameter from \(V\) through \(O\). This splits everything into two halves; look at one. In triangle \(VOA\), \(OV = OA\) (both radii), so it is isosceles and its base angles at \(V\) and \(A\) are equal; call each \(x\). The exterior angle at \(O\) of this triangle, the angle between \(\overline{OA}\) and the extension of the diameter, is the sum of the two remote interior angles: \(x + x = 2x\). The same argument on the other side gives an inscribed part \(y\) and a central part \(2y\). So the inscribed angle is \(x + y\) and the central angle is \(2x + 2y = 2(x+y)\). \(\blacksquare\) (The other cases, where the centre is outside the angle or on one side, use subtraction instead of addition.)

Consequences:

A soup can has radius 3.5 cm and height 10 cm. How much soup does it hold, and how much metal does it take to make? Why does a cone hold exactly a third of the cylinder that fits around it, not a half? And why is a sphere’s surface area exactly four times the area of its shadow?

Prisms and cylinders: stacking. A prism is a stack of identical thin slices, each with the base’s area. Stack them to height \(h\) and you get \(V = Bh\). A cylinder is a prism with a circular base, so \(V = \pi r^2 h\). This "stack of slices" idea is Cavalieri’s principle: two solids with equal cross-sectional areas at every height have equal volumes. (A stack of coins pushed sideways into a slanted pile keeps its volume.)

Cylinder surface area: unroll it. The curved side unrolls into a rectangle whose width is the circumference \(2\pi r\) and whose height is \(h\), area \(2\pi r h\). Add two circular ends: \(2\pi r^2 + 2\pi r h\).

Pyramids and cones: why one third? Here is a version you can hold in your hands. A cube of side \(s\) can be cut into three identical square pyramids, each with one face of the cube as its base and one shared corner of the cube as its apex: pick a corner, and the three faces of the cube that don’t touch it each serve as the base of one pyramid. The three pyramids are congruent and fill the cube, so each has volume \(\tfrac13 s^3 = \tfrac13 \times (\text{base}) \times (\text{height})\). Cavalieri’s principle and stretching extend this to every pyramid, and a cone is a pyramid with so many sides that its base is a circle, giving \(\tfrac13 \pi r^2 h\).

Another way to feel the third: a pyramid’s cross-section shrinks as you go up, and its area is proportional to the square of the remaining distance to the apex. Adding up slices whose areas grow like \(t^2\) gives the "one third" for the same reason that \(\int_0^1 t^2\,dt = \tfrac13\), as Chapter 19 will show.

Cone surface area. Cut the curved surface along a slant line and flatten it into a sector of a circle of radius \(\ell\) whose arc length is the base circumference \(2\pi r\). A sector’s area is proportional to its arc: it is the fraction \(\tfrac{2\pi r}{2\pi \ell}\) of the full disc \(\pi \ell^2\), giving \(\pi r \ell\).

The sphere. Archimedes considered this result his finest; he asked for it to be engraved on his tomb. Put a sphere of radius \(r\) snugly in a cylinder of radius \(r\) and height \(2r\). Then:

And the "four times its shadow" question: the sphere’s shadow is a disc of area \(\pi r^2\), and its surface is \(4\pi r^2\).

The soup can (\(r = 3.5\), \(h = 10\), in cm). Try it: Cylinder volume and area.

Since \(1 \text{ cm}^3 = 1\) ml, the can holds about 385 ml.

An ice-cream cone with radius 3 cm and height 12 cm: Try it: Cone volume.

A ball of radius 6 cm: \(V = \tfrac43 \pi (216) = 288\pi \approx 904.8 \text{ cm}^3\) and \(SA = 4\pi (36) = 144\pi \approx 452.4 \text{ cm}^2\). Try it: Sphere volume and area.

You’ve now seen several proofs in passing. What exactly makes an argument a proof? And why does the US Geometry course spend weeks having students write proofs in a rigid two-column format? The answer matters beyond geometry: a proof is a chain where each step is justified by something already accepted, which is the same discipline as reasoning about a program’s correctness, writing a loop invariant, or arguing that a distributed protocol can’t deadlock.

A proof is a conversation with a sceptic who accepts only the axioms, definitions and previously proved theorems. Each step must be one the sceptic can’t refuse. "It looks true in the diagram" is not a step. "These angles are vertical angles, so they’re equal" is.

Example 1: vertical angles are congruent.

Given: lines \(\ell\) and \(m\) intersect, forming \(\angle 1\), \(\angle 2\), \(\angle 3\) with \(\angle 1\) and \(\angle 3\) vertical.
Prove: \(\angle 1 \cong \angle 3\).

Example 2: the base angles of an isosceles triangle are equal.

Given: \(\triangle ABC\) with \(\overline{AB} \cong \overline{AC}\).
Prove: \(\angle B \cong \angle C\).

Plan (working backwards): if we can split the triangle into two congruent halves, one containing \(\angle B\) and the other \(\angle C\), CPCTC finishes the job. Draw the bisector of \(\angle A\), meeting \(\overline{BC}\) at \(D\).

As a bonus, CPCTC also gives \(BD = CD\) and \(\angle ADB = \angle ADC\); since those two angles form a linear pair, each is \(90^\circ\). So the bisector of the vertex angle is also the perpendicular bisector of the base.

Mathematicians rarely write two-column proofs. The usual form is a paragraph proof, which is the same chain of justified steps in prose, as in the angle-sum and inscribed-angle proofs earlier in this chapter. There are also flow-chart proofs (boxes connected by arrows, popular in some US textbooks) and coordinate proofs, which place the figure on axes and compute (Chapter 10). Choosing: two-column format is a training tool that makes every justification explicit, and it’s what US tests expect; paragraph proofs are what you’ll read in every other mathematics book; coordinate proofs are best when the claim is about lengths, midpoints or slopes and a clever synthetic argument isn’t obvious.