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A circle can be inscribed in quadrilateral $A B C D$. Let $K$ be the intersection point of its diagonals. It is known that $A B>B C>K C, B K=4+\sqrt{2}$, and the perimeter and area of triangle $B K C$ are 14 and 7, respectively.

Find $D C$.
Applying Heron's formula, 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 to each other, then the sums of the squares of the opposite sides are equal.

## Solution

Let $p=7$ be the semi-perimeter, $S=7$ be the area of triangle $B K C$; $L, M$ and $N$ are the points of tangency of the inscribed circle of this triangle with the sides $B C, K C$ and $B K$ respectively. Denote $K M=K N=x, B L=B N=$ $y$. Then

$$
\begin{gathered}
p-B K=7-(4+\sqrt{2})=3-\sqrt{2}, p-B C=K M=x \\
p-K C=B L=y, x+y=K N+B N=4+\sqrt{2}
\end{gathered}
$$

By Heron's formula $S=\sqrt{p(p-B K)(p-B C)(p-K C)}$, or $7=\sqrt{7(3-\sqrt{2}) x y}$. From this, we find that

$$
x y=\frac{7}{3-\sqrt{2}}=3+\sqrt{2}
$$

Since $x+y=4+\sqrt{2}$, the numbers $x$ and $y$ are the roots of the quadratic equation $t^{2}-(4+\sqrt{2}) t+3+\sqrt{2}=0$. Thus, $x=1, y=3+\sqrt{2}$ or $x=3+\sqrt{2}, y=1$.

In the first of these cases,

$$
\begin{gathered}
B C=B L+L C=B L+(p-B K)=y+7-(4+\sqrt{2})=3+\sqrt{2}+3-\sqrt{2}=6 \\
K C=K M+M C=K M+(p-B K)=x+7-(4+\sqrt{2})=1+3-\sqrt{2}=4-\sqrt{2}4-\sqrt{2}>B C
\end{gathered}
$$

which does not satisfy the condition of the problem. Therefore, $B C=6, K C=4-\sqrt{2}$.

Since

$$
B K^{2}+K C^{2}=(4+\sqrt{2})^{2}+(4-\sqrt{2})^{2}=36=B C^{2}
$$

triangle $B K C$ is a right triangle, $\angle B K C=90^{\circ}$. Thus, the diagonals of quadrilateral $A B C D$ are perpendicular.

We will prove that if the diagonals of a convex quadrilateral are perpendicular to each other, then the sums of the squares of the opposite sides are equal.

Indeed, let the diagonals $A C$ and $B D$ of such a quadrilateral $A B C D$ intersect at point $K$. By the Pythagorean theorem,

$$
A B^{2}+C D^{2}=\left(K A^{2}+K B^{2}\right)+\left(K C^{2}+K D^{2}\right)=\left(K A^{2}+K D^{2}\right)+\left(K B^{2}+K C^{2}\right)=A D^{2}+B C^{2}
$$

Let for quadrilateral $A B C D$ from our problem $A B=a, B C=b, C D=c$ and $A D=d$. Since a circle can be inscribed in it, then $a+c=b+d$. Moreover, as proven, $a^{2}+c^{2}=b^{2}+d^{2}$. Then

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ a + c = b + d } \\
{ a ^ { 2 } + c ^ { 2 } = b ^ { 2 } + d ^ { 2 } } \\
{ a > b }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a - b = d - c } \\
{ a ^ { 2 } - b ^ { 2 } = d ^ { 2 } - c ^ { 2 } } \\
{ a > b }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a-b=d-c \\
(a-b)(a+b)=(d-c)(d+c) \\
a>b
\end{array} \Leftrightarrow\right.\right.\right. \\
& \left\{\begin{array} { l } 
{ a - b = d - c } \\
{ ( a - b ) ( a + b ) = ( a - b ) ( d + c ) } \\
{ a > b }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a - b = d - c } \\
{ a + b = d + c } \\
{ a > b }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=d \\
b=c \\
a>b
\end{array}\right.\right.\right.
\end{aligned}
$$

Therefore, $C D=c=b=B C=6$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-43.jpg?height=863&width=763&top_left_y=0&top_left_x=653)

## Answer

6.