The Hinge Theorem

First follow an instance with fixed numbers. Let $AB$ have length $6$ and $AC$ length $4$, with a right angle at $A$. Inside that right angle, draw $AX$ of length $4$ at $40^\circ$ to $AB$. Triangle $ABX$ copies any triangle with arms $6$ and $4$ and an included angle of $40^\circ$, by SAS.

We accept this basic fact of plane incidence: a ray from a triangle’s vertex strictly inside its angle meets the opposite side between its endpoints.

The angle from $AX$ to $AC$ measures $50^\circ$. Bisect it with ray $AP$, making two $25^\circ$ parts. This ray meets the opposite side $BC$ at an interior point $P$.

The copied angle is $40^\circ$, while each bisected half is $25^\circ$. The equality used by SAS is between those two halves.

Triangles $ACP$ and $AXP$ have equal $4$-unit arms, the common side $AP$, and equal included angles of $25^\circ$. SAS gives

This fixed-number drawing illustrates the noncollinear Case A of the general proof below. The route from $B$ through $P$ to $X$ is longer than $BX$. Replacing $PX$ by its equal length $PC$ makes that route as long as $BC$. Thus the $90^\circ$ triangle has a longer third side than the $40^\circ$ triangle. Within this noncollinear case, the reasoning does not depend on the particular angle measures or arm lengths. The general argument below also handles collinear auxiliary points.

Now take any two triangles with the matching pairs stated above, and suppose angle $BAC$ is larger than angle $EDF$. Copy angle $EDF$ inside angle $BAC$, starting from ray $AB$. On the new ray choose $X$ so that

SAS identifies the copied triangle:

Consequently,

Bisect the positive angle $XAC$. Its bisector ray lies strictly inside angle $BAC$, so it meets segment $BC$ at a point $P$ strictly between $B$ and $C$. In particular, $BP$ and $PC$ are positive. Since $AP$ lies between $AX$ and $AC$, ray $AX$ lies strictly inside angle $BAP$.

The two equal arms $AC$ and $AX$, the common side $AP$, and the bisected angle give SAS congruence of triangles $ACP$ and $AXP$. Hence

There are two possible cases for the remaining comparison.

Case A: $B,P,X$ are noncollinear. The triangle inequality in triangle $BPX$ gives

Substitute the equal length $PC$ for $PX$. Segment addition along $BC$ then gives

Case B: $B,P,X$ are collinear. Apply the same accepted incidence fact in triangle $BAP$: ray $AX$, strictly inside angle $BAP$, meets segment $BP$ between $B$ and $P$. This meeting point is $X$: $X$ lies on both lines $AX$ and $BC$, and these distinct lines meet only once because $A$ is off $BC$. The order along the line is therefore $B,X,P,C$. These three auxiliary points do not form a triangle, so the strict triangle inequality does not apply.

Instead, segment addition gives

The last two lengths are positive, so again

Both cases give the same strict order. Finally substitute the copied third side $EF$ for $BX$:

In the $8$-versus-$12$ example, equal openings would force equal third sides by SAS. A smaller opening opposite the $12$-unit side would force it to be shorter by the hinge theorem. Both alternatives conflict with the given lengths, leaving a larger opening.

In general, retain the equal pairs $AB$ with $DE$, and $AC$ with $DF$, and suppose

If the included angles were equal, SAS would give equal third sides, contradicting the given strict order.

If angle $BAC$ were smaller than angle $EDF$, the hinge theorem, applied in the opposite order, would give

That also contradicts the given. The remaining possibility is

The numerical lengths were not needed for either contradiction, so the argument holds for every matching pair of arms.