Visual walkthrough — Quadrilaterals — square, rectangle, parallelogram, rhombus, trapezium, kite

Before anything, two plain-word ideas we will lean on the whole way.

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Step 1 — The single fact we're allowed to borrow

WHAT. We start with one thing already proven for us: in any triangle, the three inside angles add up to exactly $180°$.

WHY. A quadrilateral is one side more than a triangle. If we can somehow turn our four-sided problem into triangle-problems, we already know the answer for each triangle. So the triangle fact is the seed — we plant it and grow the rest.

PICTURE. Below, a triangle with its three corners marked $p$, $q$, $r$. The red arc reminds us: no matter how you stretch or squash it, $p+q+r=180°$.

$\underset{\text{angle at corner 1}}{\underbrace{p}}+\underset{\text{angle at corner 2}}{\underbrace{q}}+\underset{\text{angle at corner 3}}{\underbrace{r}}=\underset{\text{a straight line’s worth of turn}}{\underbrace{180°}}$

This is proven from parallel-line ideas — see Basic Geometry - Parallel Lines and Basic Geometry - Triangles. Today we use it, we don't re-prove it.

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Step 2 — Draw one diagonal and watch the shape split

WHAT. Take any quadrilateral and label its corners $A$, $B$, $C$, $D$ going around. Now draw the diagonal from $A$ to $C$.

WHY. We want triangles, because triangles are the only shape whose angle-sum we know. A diagonal is the cheapest way to cut a four-corner shape into triangles — one straight line does it.

PICTURE. The red line $AC$ slices the quadrilateral cleanly into two triangles: $△ABC$ on one side, $△ACD$ on the other. Nothing spills outside; nothing is left over.

Notice: the diagonal touches only two of the four corners ($A$ and $C$). Corners $B$ and $D$ are left whole — each sits entirely inside one triangle. Hold that thought; it's the key to Step 4.

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Step 3 — Add up the two triangles' angles

WHAT. Each triangle's angles sum to $180°$. We have two triangles, so the total of all six little angles is:

$\underset{△ABC}{\underbrace{180°}}+\underset{△ACD}{\underbrace{180°}}=360°$

WHY. We're not doing anything clever yet — just adding what Step 1 guarantees for each half. Two halves, $180°$ each.

PICTURE. The figure shades the two triangles differently and marks all six angles: three in the upper triangle, three in the lower. The red total banner reads $360°$.

But wait — these are six small angles. The quadrilateral only has four corners. Are these the same $360°$ we care about? That's exactly what Step 4 checks.

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Step 4 — Show the six small angles rebuild the four real corners

WHAT. We prove that adding the six triangle-angles is the same as adding the four corner angles of the quadrilateral.

WHY. This is the make-or-break step. If the diagonal chopped a corner into pieces that don't fully rebuild it, our sum would be wrong. So we account for every angle honestly.

PICTURE. Look at each corner in turn.

• Corner $B$: the diagonal never touched it. Its whole angle $\angle B$ lives inside $△ABC$, untouched. ✓
• Corner $D$: same story — the whole angle $\angle D$ lives inside $△ACD$, untouched. ✓
• Corner $A$: the diagonal did land here. It split $\angle A$ into two slices — one slice in each triangle. Add the slices back and you get the whole corner: $\underset{\text{true corner}}{\underbrace{\angle A}}=\underset{\text{slice in }△ABC}{\underbrace{\angle BAC}}+\underset{\text{slice in }△ACD}{\underbrace{\angle CAD}}$
• Corner $C$: likewise split into two slices that add back to $\angle C$.

So the six triangle-angles are exactly: whole $\angle B$, whole $\angle D$, the two slices of $\angle A$, and the two slices of $\angle C$. Reassembling the slices, they add up to precisely

$\angle A+\angle B+\angle C+\angle D.$

Nothing added, nothing lost — the red arrows show each slice sliding home into its true corner.

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Step 5 — Snap the two results together

WHAT. Step 3 said the six angles total $360°$. Step 4 said the six angles equal $\angle A+\angle B+\angle C+\angle D$. Set them equal:

$\underset{\text{the four real corners}}{\underbrace{\angle A+\angle B+\angle C+\angle D}}=\underset{\text{two triangles}}{\underbrace{360°}}$

WHY. Two true statements about the same six angles must agree. That agreement is our theorem.

PICTURE. The four corner arcs, each a different shade, drawn onto the whole quadrilateral, with the red $=360°$ result underneath.

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Step 6 — Does it survive weird shapes? (edge cases)

WHAT. We test the proof on shapes that try to break it.

WHY. A proof you only checked on a neat rectangle isn't trusted yet. The reader must never meet a case we skipped.

PICTURE. Three troublemakers, side by side.

• A long thin parallelogram (case 1): The two triangles are skinny, but each still has angle-sum $180°$. Steps 3–5 never assumed anything about size or slant. ✓
• A dented (concave) quadrilateral (case 2): One corner points inward, past $180°$. Here you must draw the diagonal that stays inside the shape (the red one). It still makes two triangles, still $360°$. The reflex-looking corner is genuinely a big interior angle, and the sum still lands on $360°$. ✓
• A near-flattened quadrilateral (case 3, the limit): Push one corner until it almost lies on the opposite side. One angle shrinks toward $0°$ and another swells toward the slack — but their sum with the other two is still forced to $360°$. Nothing special happens; the formula holds right up to the degenerate edge. ✓

The only rule: pick the diagonal that lies inside the shape. For a dented quadrilateral one of the two diagonals goes outside — don't use that one, because then the "triangles" would overlap and double-count.

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The one-picture summary

One diagonal → two triangles → $180°+180°=360°$ → and those six angles reassemble into the four true corners. That single chain is the entire proof.

Recall Feynman retelling — say it out loud in plain words

"I've got a four-cornered shape and I want the four corner angles to add up. I don't know a rule for four-cornered shapes, but I do know one for triangles — their angles add to a straight-line $180°$. So I draw a shortcut line across the middle, from one corner to the opposite corner. That single line cuts my shape into two triangles. Each triangle gives me $180°$, so together they give $360°$. Now the sneaky part: are those triangle-angles the same as my real corners? Two of my corners were never touched by the line, so they're fine as they are. The other two corners got sliced in half by the line — but if I glue the halves back, I get the whole corner again. So the six triangle-angles are exactly my four corners, no more, no less. Therefore my four corners add to $360°$. And this never breaks — thin, dented, or nearly flat — as long as I draw the diagonal that stays inside."

Recall Quick checks

Why do a triangle's angles matter here? ::: A quadrilateral splits into two triangles, and each triangle's angles sum to $180°$. How many triangles does one diagonal create? ::: Exactly two. Why isn't the answer just "two triangles so more than 360"? ::: The six triangle-angles ARE the four corners (two corners get split into slices that add back), so the total is exactly $360°$. Which diagonal must you pick for a dented (concave) quadrilateral? ::: The one that lies entirely inside the shape, so the two triangles don't overlap. What happens to the sum as the shape flattens to a degenerate line? ::: It stays $360°$; individual angles shift, but the total is fixed.

Where this leads: the parallelogram's "opposite angles equal" and "consecutive angles supplementary" facts (see the parent Quadrilaterals — square, rectangle, parallelogram, rhombus, trapezium, kite) both start from this $360°$. Area work builds on Mensuration - Area and Perimeter, side-length work on Basic Geometry - Pythagorean Theorem and Coordinate Geometry - Distance Formula.