Half-angle model (rotational stitching)

When to use

• A $45°$ angle inside a square (the classic): in square $ABCD$, vertex $A$ opens an angle $\angle EAF = 45°$, with $E$ on $BC$ and $F$ on $CD$
• A $30°$ angle inside an equilateral triangle: equilateral $\triangle ABC$, vertex $A$ opens $\angle EAF = 30°$
• More generally: an isosceles figure with apex angle $2\alpha$ exposes a single $\alpha$ angle ("half the apex") at the vertex
• You need to show one segment equals the sum of two others (e.g. $EF = BE + DF$)

Core move

Rotate one piece around the apex by $2\alpha$ — the three pieces stitch into two congruent triangles, and the two polyline segments line up into a single straight segment in their new position.

Construction

Take the square $ABCD$ + $\angle EAF = 45°$ ($E$ on $BC$, $F$ on $CD$):

1. Rotate one piece: rotate $\triangle ABE$ about $A$ by $90°$ clockwise ($=$ apex $2\alpha$): $B \to D$, $E \to E'$.

   

2. New points are collinear: $\angle ADE' = \angle ABE = 90°$ and $\angle ADF = 90°$, so $E'$, $D$, $F$ are collinear.

3. The half-angle reappears: $\angle E'AF = \angle E'AD + \angle DAF = \angle BAE + \angle DAF = 90° - 45° = 45° = \angle EAF$.

4. SAS congruence: $AE' = AE$ (rotation), $AF = AF$ (shared), $\angle E'AF = \angle EAF$, so $\triangle AE'F \cong \triangle AEF$.

   

5. Read the conclusion: $EF = E'F = E'D + DF = BE + DF$.

Why it works

After rotating $B$ to $D$, the only "angle structure around $A$" left is one $\alpha$ (the half-angle) and one $\alpha$ (the $E'$-to-$F$ angle after rotation) — precisely the side-angle-side ingredients for SAS congruence. The essence: half-apex + apex-rotation = the bend in the polyline disappears.

Worked examples

• Square $ABCD$, $\angle EAF = 45°$, $E \in BC$, $F \in CD$: show $EF = BE + DF$
• Square $ABCD$, $P$ on the diagonal, $\angle MPN = 90°$, $M \in AB$, $N \in BC$: find the extrema of $PM \cdot PN$

Variants / generalizations

• Equilateral triangle + $30°$ half-angle: apex $60°$, half-angle $30°$, rotate $60°$ to stitch
• General isosceles + half-apex: apex $2\alpha$, half $\alpha$, rotate $2\alpha$; the conclusion "$EF = BE + DF$" upgrades to an equality scaled by the leg-length coefficient
• Non-isosceles apex: the isosceles condition $AB = AD$ is mandatory (rotation preserves distance); without it you must use the similarity version (Melon-and-bean (spiral similarity)).
• The $\sqrt 2$ (square) and $\sqrt 3$ (equilateral) coefficients tied to half-angle $\alpha$ are essentially $2\sin\alpha$ — a hint that they share the same mechanism as the weight triangle in Weighted Fermat Point (rotation + scaling lemma).