Pythagorean theorem — proof (by similar triangles, rearrangement), converse: Visual walkthrough

This page rebuilds the rearrangement proof of the Pythagorean theorem from absolute zero. No formula is used before we can see it. By the end you will have watched $a^{2} + b^{2} = c^{2}$ assemble itself out of tiles, and you will know why the right angle is the secret ingredient.

Step 0 — The three words we will use

Before any picture, three plain-word definitions. We use no symbol until it is anchored here.

In the figure: the little box marks the right angle. The coral side has length $a$ , the mint side has length $b$ , and the lavender slanted side has length $c$ . The two small arcs mark the acute angles $α$ and $β$ . Everything from here on is built on this one triangle.

Why these names matter

We will build squares on

a

, on

b

, and on

c

; the whole proof is comparing their areas.

Step 1 — Turn a length into an area

WHAT. Take the length $c$ and build a square on it: a shape with all four sides equal to $c$ .

WHY. The theorem is a statement about area, not length. A length like $c$ is a one-dimensional thing; to add it to other lengths in a meaningful area-conserving way we must square it. "Squaring" literally means make a square.

PICTURE.

So the goal $a^{2} + b^{2} = c^{2}$ now reads in plain English: the mint square plus the coral square equals the lavender square. Let us make that happen with tiles.

Step 2 — Build a big frame that holds four triangles

WHAT. Draw one large square whose side length is $a + b$ (one leg laid end-to-end with the other leg). Call this whole filled-in $(a + b) \times (a + b)$ square region the frame. Its total area is $(a + b) \times (a + b)$ .

WHY. A frame of side $a + b$ is exactly wide enough to seat four copies of our triangle around its rim without overlap. That "just fits" is the engine of the proof — it lets us fill one region two different ways.

PICTURE.

This expansion is just the area of the whole frame, sliced into four pieces. Hold onto it — it is our first way of measuring the frame.

Step 3 — Seat the four triangles inside the frame

WHAT. Place four identical copies of our right triangle inside the frame, each in a corner, hypotenuses pointing inward.

WHY. Each triangle has area $\frac{1}{2} ab$ (half of a leg-by-leg rectangle). Four of them together cover part of the frame. Whatever is left over in the middle is the region we care about.

PICTURE.

Notice the four hypotenuses (lavender) now enclose a tilted shape in the very middle. That middle is the star of Step 4.

Step 4 — Prove the middle hole is a genuine square

WHAT. Show the tilted region in the centre has four equal sides (each length $c$ ) and four right-angle corners — so it is a true square of area $c^{2}$ .

WHY. If it is a square of side $c$ , its area is $c^{2}$ , and we can see $c^{2}$ sitting inside the frame. This is where the right angle does its secret work.

PICTURE.

Each side of the middle hole is a hypotenuse, so all four sides equal $c$ — equal-length sides, good. But why are its corners right angles?

Recall from Step 0 that $α$ and $β$ are the two acute angles of our triangle. Look at any point on the frame's edge where two triangles meet (marked on the figure). Three angles sit along that straight edge: the angle $α$ from one triangle, the corner of the hole, and the angle $β$ from the neighbouring triangle — the two triangles present different acute angles at that point because they are rotated a quarter-turn from each other.

Four equal sides and four right angles ⇒ the hole is a square. Its area is $c^{2}$ . This is the only step that uses the right angle — and without it the hole would be a slanted diamond, not a square, and the whole proof collapses.

Step 5 — Measure the frame the second way

WHAT. Add up the pieces we can now see inside the frame (again, the whole $(a + b) \times (a + b)$ square): four triangles plus the middle square.

WHY. This is our second way of measuring the exact same frame from Step 2. Two honest measurements of one area must agree.

PICTURE.

Step 6 — Set the two measurements equal and cancel

WHAT. The frame's area from Step 2 and from Step 5 describe the same square, so equate them, then remove what is common.

WHY. Anything measured two correct ways gives the same number. Subtracting the shared $2 ab$ leaves the theorem standing alone.

PICTURE.

Step 7 — The degenerate cases (never leave a gap)

WHAT. Check what happens when a leg shrinks to nothing, or the two legs are equal.

WHY. A proof you trust must survive its extreme inputs, not just the pretty ones.

PICTURE.

A leg is zero ( $a = 0$ ). The triangle flattens into a line segment; the hypotenuse lies right on top of leg $b$ . The formula gives $0^{2} + b^{2} = b^{2}$ , so $c = b$ . True — the "triangle" is just a stick of length $b$ . The formula degrades gracefully.
The other leg is zero ( $b = 0$ ). By the exact same argument with the roles swapped, $a^{2} + 0^{2} = a^{2}$ , so $c = a$ . The triangle collapses onto leg $a$ . Both legs are on equal footing, so either one vanishing gives the same graceful behaviour — we call out both so no reader wonders about the symmetric case.
Equal legs ( $a = b$ ). The triangle is a half-square (an isosceles right triangle). Then $c^{2} = a^{2} + a^{2} = 2 a^{2}$ , so $c = a 2$ . The hypotenuse is $2 \approx 1.414$ times a leg — the same diagonal that appears in every square tile.
What if the corner is not a right angle? Step 4 fails — the middle hole becomes a slanted diamond, its area is no longer $c^{2}$ , and equality breaks. That is exactly why $a^{2} + b^{2} = c^{2}$ is a right-triangle-only statement, and why for other triangles you must reach for the Law of Cosines.

The one-picture summary

One frame of side $a + b$ , filled two ways: four triangles + a $c^{2}$ square on the left, and four triangles + an $a^{2}$ square + a $b^{2}$ square on the right. Peel off the four triangles from each side and the surviving squares must match: $a^{2} + b^{2} = c^{2}$ .

Recall Feynman retelling — say it like you would to a friend

Take your right triangle and make four photocopies. Build a big square whose sides are as long as one short side plus the other short side. Now shove the four triangles into the four corners. Two things can happen depending on how you nudge them — but the empty space left over is always the same amount, because you never added or removed any triangle.

Arrange them one way and the leftover hole is a single tilted square whose side is the long slanted side, $c$ — its area is $c^{2}$ . Slide the same four triangles into a different pattern and the leftover splits into two upright squares, one of side $a$ and one of side $b$ — areas $a^{2}$ and $b^{2}$ . Same triangles, same big frame, same leftover amount of space. So $c^{2}$ and $a^{2} + b^{2}$ are just two names for the identical leftover: $a^{2} + b^{2} = c^{2}$ .

The only place the right angle sneaks in is when we claim the tilted hole is a clean square. Two non-right angles of a right triangle always add to $9 0^{\circ}$ , which forces the hole's corner to be $9 0^{\circ}$ too. Bend the corner away from square and the hole becomes a lopsided diamond — the trick dies, and that is precisely why the rule only works for right triangles.

Recall Quick self-test

Why must a leg be squared and not just added? ::: The theorem conserves area; a length must become a square (a 2-D area) before areas can be compared and added. Where is the right angle actually used in the whole proof? ::: Only in Step 4, to prove the middle hole is a true square of area $c^{2}$ . What is $c$ when $a = b = 1$ ? ::: $c = 2$ , since $c^{2} = 1 + 1 = 2$ . If $a^{2} + b^{2} \neq = c^{2}$ , what does that tell you about the triangle? ::: It is not a right triangle — use the Law of Cosines instead.

Where this leads: the same area-conservation idea powers the distance formula, generalises to the 3D Distance Formula, measures the length of vectors, gives the modulus of Complex Numbers, and its whole-number solutions are the Pythagorean Triples.