[ $\left.\begin{array}{l}\text { Inscribed quadrilaterals } \\ \text { [Quadrilateral: calculations, metric relations.] }\end{array}\right]$

In the quadrilateral $A B C D$, a circle can be inscribed. Let $K$ be the point of intersection of its diagonals. It is known that $B C>A B>B K, K C=\sqrt{7}-1$, the cosine of angle $K B C$ is $\frac{\sqrt{7}+1}{4}$, and the perimeter of triangle $B K C$ is $2 \sqrt{7}+4$. Find $D C$.
Let $\angle K B C=\alpha, B K=x, B C=y$. Let $P=2 \sqrt{7}+4$ be the perimeter of triangle $B K C$. Then

$$
x+y=P-K C=2 \sqrt{7}+4-(\sqrt{7}-1)=5+\sqrt{7}
$$

By the cosine rule,

$$
K C^{2}=B K^{2}+B C^{2}-2 \cdot B K \cdot B C \cdot \cos \alpha, \text { or }(\sqrt{7}-1)^{2}=x^{2}+y^{2}-2 x y \cdot \frac{\sqrt{7}+1}{4} .
$$

Since

$$
x^{2}+y^{2}=(x+y)^{2}-2 x y=(5+\sqrt{7})^{2}-2 x y=32+10 \sqrt{7}-2 x y
$$

from the second of the obtained equations, we find that

$$
x y=\frac{24(2+\sqrt{7})}{5+\sqrt{7}}=4(1+\sqrt{7})
$$

From the system

$$
\left\{\begin{array}{l}
x+y=5+\sqrt{7} \\
x y=4(1+\sqrt{7}) \\
x a }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a - y = d - c } \\
{ a ^ { 2 } - y ^ { 2 } = d ^ { 2 } - c ^ { 2 } } \\
{ y > a }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a-y=d-c \\
(a-y)(a+y)=(d-c)(d+c) \\
y>a
\end{array} \Leftrightarrow\right.\right.\right. \\
& \left\{\begin{array}{l}
a-y=d-c \\
(a-y)(a+y) \\
y>a
\end{array}=(a-y)(d+c) \Leftrightarrow\left\{\begin{array} { l } 
{ a - y = d - c } \\
{ a + y = d + c } \\
{ y > a }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=d \\
y=c \\
y>a
\end{array}\right.\right.\right.
\end{aligned}
$$

Therefore, $C D=c=y=B C=4$.

Using the cosine rule, find the sides of triangle $B K C$. Prove that $A C \perp B D$. Also prove that if the diagonals of a convex quadrilateral are perpendicular, then the sums of the squares of the opposite sides are equal.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-44.jpg?height=612&width=571&top_left_y=1512&top_left_x=755)

## Answer

4.