Saudi Arabian Mathematical Competitions, 2012

Let $M$ be the intersection point of the diagonals $AC$ and $BD$.

Since $\hat{B} \ne 90^\circ$ it follows that $\hat{D} \ne 90^\circ$, and hence we have $\cos \hat{B} \ne 0$ and $\cos \hat{D} \ne 0$. The quadrilateral $ABCD$ is cyclic, so therefore $\hat{B} + \hat{D} = 180^\circ$. It follows $\cos \hat{B} + \cos \hat{D} = 0$, and by the Cosine Law we obtain
$$\frac{AB^2 + BC^2 - AC^2}{2AB \cdot BC} + \frac{AD^2 + DC^2 - AC^2}{2AD \cdot DC} = 0. \quad (1)$$
The given relation is equivalent to
$$AB^2 + BC^2 - AC^2 = -(AD^2 + DC^2 - AC^2) \neq 0,$$
and from (1) it follows $AB \cdot BC = AD \cdot DC$. This implies that $K[ABC] = K[ADC]$, and hence $BB' = DD'$, where $BB'$ and $DD'$ are the altitudes of triangles $ABC$ and $ADC$, respectively.

The triangles $BB'M$ and $DD'M$ are congruent, so $MB = MD$, and we are done.