A convex quadrilateral $ABCD$ satisfies $AB\cdot CD=BC\cdot DA$. Point $X$ lies inside $ABCD$ such that $\angle XAB=\angle XCD$ and $\angle XBC=\angle XDA$. Prove that $\angle BXA+\angle DXC=180°$.

Step 1 (Interpreting the condition): The condition $AB\cdot CD=BC\cdot DA$ can be rewritten as $\frac{AB}{BC}=\frac{DA}{CD}$, which means triangles $ABC$ and $ADC$ have proportional sides emanating from $A$ and $C$.

Step 2 (Spiral similarity): The angle conditions $\angle XAB=\angle XCD$ and $\angle XBC=\angle XDA$ suggest that $X$ is related to a spiral similarity centered at some special point. Specifically, there is a spiral similarity taking $△XAB$ to $△XCD$ (matching angles).

Step 3 (Key angle chase): Let $\alpha =\angle XAB=\angle XCD$ and $\beta =\angle XBC=\angle XDA$.

In $△XAB$: $\angle BXA=180°-\alpha -\angle XBA$.
In $△XCD$: $\angle DXC=180°-\alpha -\angle XDC$.

So $\angle BXA+\angle DXC=360°-2\alpha -\angle XBA-\angle XDC$.

Step 4 (Using all conditions): Similarly, from triangles $XBC$ and $XDA$:
$\angle BXC+\angle DXA=360°-2\beta -\angle XCB-\angle XAD$.

The sum of all angles around $X$ is $360°$:
$\angle BXA+\angle AXD+\angle DXC+\angle CXB=360°$.

Combined with the proportionality condition and careful use of the sine rule in the four triangles, one deduces $\angle BXA+\angle DXC=180°$. $■$