$:$ Folklore

The center $O$ of the circle circumscribed around the quadrilateral $ABCD$ lies inside it. Find the area of the quadrilateral if $\angle BAO = \angle DAC$, $AC = m$, $BD = n$.
Let the diagonals $ABCD$ intersect at point $P$. We will prove that $AC$ and $BD$ are perpendicular.

First method. We use the obvious fact: if $AP$ is the altitude of triangle $ABC$, and $O$ is the center of its circumscribed circle, then $\angle OAB = \angle PAC$ (see the left figure).

Applying this fact to triangle $ABD$ (see the right figure), we get that from the condition $\angle BAO = \angle DAC$ and the uniqueness of the perpendicular dropped from point $A$ to line $BD$, it follows that $AP \perp BD$.
![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-44.jpg?height=664&width=1100&top_left_y=336&top_left_x=474)

Second method. Draw the diameter $AA'$, then from the equality $\angle BAO = \angle DAC$ it follows that $\angle BAP = \angle DAA'$ (see figure b). Moreover, $\angle ABP = \angle AA'D$ (inscribed angles subtending the same arc). Since the inscribed angle $ADA'$ subtends the diameter, $\angle ADA' = 90^\circ$, hence,

$\angle BAP + \angle ABP = \angle DAA' + \angle AA'D = 90^\circ$. Thus, the angle $APB$ between the diagonals of the quadrilateral is a right angle.

Therefore, $S_{ABCD} = \frac{1}{2} AC \cdot BD = \frac{mn}{2}$.

## Answer

$\frac{mn}{2}$.