In quadrilateral $A B C D$, it is known that $D O=4, B C=5, \angle A B D=45^{\circ}$, where $O$ is the point of intersection of the diagonals. Find $B O$, if the area of quadrilateral $A B C D$ is equal to $\frac{1}{2}(A B \cdot C D+B C \cdot A D)$.
Prove that $ABCD$ is a cyclic quadrilateral with mutually perpendicular diagonals.

## Solution

Let $C_{1}$ be the point symmetric to vertex $C$ with respect to the perpendicular bisector of diagonal $BD$. Then

$$
\begin{gathered}
S_{\triangle \mathrm{ABCD}}=S_{\Delta \mathrm{ABC}_{1} \mathrm{D}}=S_{\Delta \mathrm{ABC}_{1}}+S_{\Delta \mathrm{AC}_{1} \mathrm{D}}= \\
=\frac{1}{2} \cdot A B \cdot B C_{1} \sin \angle A B C_{1}+\frac{1}{2} \cdot A D \cdot D C_{1} \sin \angle A D C_{1} \leqslant \\
\leqslant \frac{1}{2}\left(A B \cdot B C_{1}+A D \cdot D C_{1}\right)=\frac{1}{2}(A B \cdot D C+A D \cdot B C)
\end{gathered}
$$

with equality only when

$$
\angle A B C_{1}=\angle A D C_{1}=90^{\circ}
$$

Therefore, quadrilateral $A B C_{1} D$ is cyclic and $A C_{1}$ is the diameter of its circumscribed circle.

The perpendicular bisector of diagonal $BD$ is the axis of symmetry of this circle. Therefore, vertex $C$ also lies on the circle. Hence, quadrilateral $A B C D$ is cyclic. Since $A C_{1}$ is the diameter of its circumscribed circle, $\angle A C C_{1}=90^{\circ}$. Therefore, diagonal $A C$ is parallel to the perpendicular bisector of diagonal $B D$. Consequently, diagonals $A C$ and $B D$ of quadrilateral $A B C D$ are mutually perpendicular. Then in triangle $COD$:

$$
O D=4, \angle C O D=90^{\circ}, \angle O C D=\angle A C D=\angle A B D=45^{\circ}
$$

Therefore, $O C=O D=4$. By the Pythagorean theorem in right triangle $COB$, we find that

$$
O B=\sqrt{B C^{2}-O C^{2}}=\sqrt{5^{2}-4^{2}}=3
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-33.jpg?height=475&width=558&top_left_y=623&top_left_x=750)

## Answer

3.