Условие:
$N 25$.
Dано: $A B C N-$ таи. $A B \perp B C$ O+ T. $C, D, \varepsilon-$ на oup., т. $\varepsilon \in A B$ $A D=6$
$B C=5$
Haĩmu: \& $H-$ ?
CN. NUCT 4
3. m.l. $A B C D$-тpan. $\Rightarrow B C \| A M$, no ycu. $A B \perp B C$, brac. $C M \perp B C$,
$\Rightarrow C M 1 \| A B, \angle A B C=90^{\circ}, \angle B A M=90^{\circ}$ (ho $C b-B y$ тpan.), $C M \perp A D$
$=C N A=90^{\circ}, \angle B C N=90^{\circ}(C M-6 M C ., \angle M \perp B C, C M \perp A D)$
4. $\quad C N \| A B(n .3)$
$B C / / A M(n .3)$
5. $A D=A M+M D \Rightarrow M D=A D-A M=A D-B C=6-5=1$
6. paccroтpur $\triangle K B C$ и $\triangle C M D$ :
$
\begin{array}{l}
4 \\
\frac{B C}{M D}=\frac{K C}{C D}
\end{array}
$
$
\begin{array}{l}
\varepsilon K^{2}=K D \cdot K C, K D=\left(K C+C D \Rightarrow \quad \Rightarrow \quad \frac{5}{1}=\frac{K C}{C D} \Rightarrow \begin{array}{l}
K C= \\
\varepsilon K^{2}=K D D
\end{array}\right.
\end{array}
$
T.K. $K C=5 C D \Rightarrow E K^{2}=6 C D \cdot C D \Rightarrow \varepsilon K=\sqrt{6 C D^{2}}=C D \cdot \sqrt{6}$
$
\begin{array}{l}
\therefore \angle K B C=\angle C N O=90^{\circ}
\end{array}
$