Some theorems are true because you can see that they must be. No symbol-pushing, no induction — just shapes that rearrange, and an area that refuses to change. Each tab below is a classic "proof without words" you can drag, slide and reassemble: the pieces move, a running area ✓ readout confirms nothing was created or destroyed, and the identity falls out of the picture. Everything is computed live — the counters read the geometry, they are not captions.
Both homes live inside the same bounding square of side a+b, whose area is fixed at (a+b)² = a² + 2ab + b². And both homes contain the same four right triangles, each with legs a and b, so each triangle has area ½ab and the four together always cover 2ab.
Subtraction does the rest. The white area not covered by triangles is (a+b)² − 2ab = a² + b² — no matter how the triangles are arranged. In the first home that white area is honestly drawn as an a×a square beside a b×b square. In the second home it is a single square whose side is the triangles' hypotenuse c.
Two ways of measuring the identical leftover must agree, so a² + b² = c². The dragging matters: because the triangles are rigid (congruent throughout), you are literally watching a conservation law — area is invariant under sliding pieces around.
This is essentially the arrangement in Zhoubi Suanjing (China, ~1st c. BCE) and the dissection favoured by Bhāskara, whose entire "proof" was reputedly the single word "Behold!" beneath the figure.
The converse is visible too: if you drag to a case where the central quadrilateral is not a square (it never is here, because all four triangles are congruent right triangles), the tiling would fail — the right angle is exactly what makes the inner shape close up into a square of side c.
Caution about "proofs without words": the picture convinces because the pieces are provably congruent and the frame is provably a square. A sloppy dissection can appear to prove false things (see the "missing square" puzzle). The honesty here is that every length is computed, not drawn by eye.
Start with a 1×1 square. To turn a k×k square into a (k+1)×(k+1) square you add cells along one row and one column, sharing the corner: that's k + k + 1 = 2k+1 cells. As k runs 0,1,2,… the borders are 1, 3, 5, 7, … — precisely the odd numbers.
So building an n×n square from the corner outward peels apart as 1 + 3 + 5 + ⋯ + (2n−1) = n². Nothing is approximated; each odd number is a physical rim, and the rims tile the square with no gaps or overlaps.
The Greek word gnomon originally meant the pointer of a sundial, then the carpenter's L-square, then exactly this L-shaped figure. The Pythagoreans arranged pebbles ("psephoi") this way and read arithmetic off the shapes — the first "figurate numbers."
The same idea keeps giving: the difference between consecutive squares, (k+1)² − k² = 2k+1, is the gnomon count, which is why the running square-differences are the odd numbers. Slide k and watch the LAST ODD ADDED card equal the jump in the square total.
Column i of the staircase is i blocks tall, so the staircase holds 1+2+⋯+n blocks. Rotating a copy by 180° turns "short next to tall" into "tall next to short," and the two profiles are complementary: every column of the combined figure is exactly n+1 tall. With n columns, that's an n × (n+1) rectangle.
Because the rectangle is two identical staircases, 2·Tₙ = n(n+1), hence Tₙ = n(n+1)/2.
The algebraic twin of this picture is Gauss's famous pairing: write 1+2+⋯+n forwards and backwards, add column-wise, and every pair sums to n+1. There are n pairs, giving n(n+1), then halve. Same theorem, same "double it and halve."
Triangular numbers Tₙ show up everywhere — handshakes among n people, edges in a complete graph, the "n choose 2" count. The picture is the reason all of those equal n(n+1)/2.
After k tiles of ratio r, the coloured fraction is Sₖ = r + r² + ⋯ + rᵏ and the uncoloured leftover is exactly rᵏ (for r = ½ that's the vanishing corner square). Multiply the leftover by (1−r) each new term and the telescoping is visible: Sₖ = (r − r^{k+1})/(1 − r).
As k → ∞ with 0 < r < 1, the leftover rᵏ → 0, so the sum limits to r/(1 − r). For r = ½ that's 1 — the whole square. The picture makes convergence visible: you can literally see there's nothing left to fill.
This dissolves Zeno's paradox. To cross a room you first cross half, then half the rest, forever — infinitely many steps. Zeno concluded motion was impossible. But the step sizes form exactly ½+¼+⅛+⋯, whose total is a finite 1: infinitely many pieces, finite distance, finite time.
The condition |r| < 1 is essential. With r ≥ 1 the tiles don't shrink and the leftover never vanishes — the series diverges, and no picture inside a bounded square could hold it.
Put a and b end to end as a diameter of length a+b; the radius is (a+b)/2 — the arithmetic mean, AM. Any point on the semicircle sees the diameter at a right angle (Thales' theorem). Drop the altitude of that right triangle to the diameter at the split point: the geometric-mean relation for a right triangle's altitude gives its length as √(a·b) — the geometric mean, GM.
That altitude is a half-chord from the diameter up to the arc. It is longest when it sits at the centre, where it equals the radius; anywhere else it's strictly shorter. Hence √(ab) ≤ (a+b)/2, with equality iff the split is central, i.e. a = b.
The equivalent "rectangle vs. square of equal perimeter" story is the same fact: among all rectangles with a fixed perimeter (fixed a+b), the square (a = b) has the largest area (ab), and its side is the AM while √(area) is the GM. Elongate the rectangle and area falls away — exactly the chord shortening.
AM–GM generalises to n numbers and underpins huge swathes of optimization and inequality theory, but the two-variable case is this one picture. It's also why "spread lowers the product for a fixed sum" — a fact that quietly drives results from finance to information theory.