Why the Formulas Work

Draw the line through one vertex PARALLEL to the opposite side. The two new angles at that vertex equal the two far angles of the triangle (alternate interior angles), and together with the vertex angle they fill a straight line: $A + B + C = 180^\circ$. $\blacksquare$

In plain terms. Slide the two bottom corners up to the top corner along parallel lines — the three angles line up along a ruler.

Example. Angles $50^\circ$ and $60^\circ$ force the third to be $180 - 110 = 70^\circ$.

Pick one vertex and draw all diagonals from it. They cut the polygon into exactly $n - 2$ triangles, and the polygon’s interior angles are exactly the triangles’ angles combined: $(n-2)\cdot 180^\circ$. $\blacksquare$

In plain terms. Fan the shape into triangles from one corner and count them: always two fewer than the sides.

Example. A hexagon fans into $4$ triangles: $4 \cdot 180^\circ = 720^\circ$.

Take a big square of side $a + b$ and tuck four copies of the right triangle into its corners. What remains uncovered is a tilted square on the hypotenuse: area $(a+b)^2 - 4\cdot\tfrac12 ab = c^2$, and expanding the left side gives $a^2 + b^2 = c^2$. $\blacksquare$

In plain terms. Four copies of the same triangle leave a $c$-square of leftover space — and computing the leftover two ways forces the theorem.

Example. Legs $3$ and $4$: $9 + 16 = 25$, so the hypotenuse is $5$.

Write the sum forwards and backwards and add columns: every column totals $n + 1$, and there are $n$ columns, so TWICE the sum is $n(n+1)$. $\blacksquare$

In plain terms. Pair the smallest with the largest, second-smallest with second-largest — every pair is worth the same.

Example. $1 + 2 + \cdots + 100$: fifty pairs of $101$ gives $5050$.