4.1. In a circle, two perpendicular chords $A B$ and $C D$ are drawn. Determine the distance between the midpoint of segment $A D$ and the line $B C$, if $A C=6, B C=5, B D=3$. Round the answer to two decimal places if necessary.
Answer: 4.24 (exact value: $\sqrt{5}+2$).

Solution. Let $AC = a$, $BC = b$, $BD = c$, $N$ be the point of intersection of the chords, $M$ be the midpoint of $AD$, and $H$ be the point of intersection of the lines $MN$ and $BC$. Denote $\angle BAD = \alpha$, $\angle CDA = \beta$, $\alpha + \beta = 90^\circ$. Then $\angle BNH = \angle ANM = \angle MAN = \alpha$, $\angle CNH = \angle MND = \angle NDM = \beta$, $\angle BCD = \angle DAN = \alpha$, $\angle CBA = \angle CDA = \beta$ as angles subtending the same chord. Therefore, the lines $NH$ and $BC$ are perpendicular, and the desired distance is the length of the segment $MH$.

Let $AN = x$, $NB = y$, $CN = u$, $ND = v$. From the similarity of $\triangle ACN$ and $\triangle DBN$ and the Pythagorean theorem, we get $\frac{u}{y} = \frac{a}{c}$, $u^2 + y^2 = b^2$, $x^2 = a^2 - u^2$, $v^2 = c^2 - y^2$. In the end,

$$
MH = \frac{\sqrt{x^2 + v^2}}{2} + \frac{uy}{b} = \frac{\sqrt{a^2 + c^2 - b^2}}{2} + \frac{abc}{a^2 + c^2}
$$

which for $AC = 6$, $BC = 5$, $BD = 3$ gives $MH = \sqrt{5} + 2 \approx 4.24$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c22e465a9d4d7e156a81g-2.jpg?height=448&width=463&top_left_y=1963&top_left_x=1476)