Let $ABCD$ be an inscribed (cyclic) quadrilateral with $\widehat{BAC}$ $\equiv $ $\widehat{DAC}$. Prove that the area of $ABCD$ is equal to $\dfrac{1}{2}AC^2\sin A$.
https://www.geogebra.org/geometry/pqkg5fgx
It's relatively easy to prove that those angles I've highlighted are equal. In addition, if you apply the area formula for the two triangles, you get $$AC^2 = AB \cdot AD + BC \cdot CD$$ (or $BC^2$). From here I've tried using similar triangles such as $ABC$ and $AQD$, where $Q$ is the intersection of the diagonals, the bisector theorem and the power of the point but it doesn't work.
Let $\measuredangle BCA=x$, $\measuredangle BAD=\alpha$, $AC\cap BD=\{K\}$ and $R$ be a radius of the circle.
Thus, $$S_{ABCD}=\frac{1}{2}AC\cdot BD\sin\measuredangle CKD=\frac{1}{2}AC\cdot 2R\sin\alpha\sin\left(x+\frac{\alpha}{2}\right)=$$ $$=\frac{1}{2}AC\cdot2R\sin\left(180^{\circ}-x-\frac{\alpha}{2}\right)\sin\alpha=\frac{1}{2}AC^2\sin\alpha.$$