8.3. Determine the real numbers $a_{1}, a_{2}, a_{3}, a_{4}$, which satisfy the equality

$$
\sqrt{a_{1}}+\sqrt{a_{2}-1}+\sqrt{a_{3}-2}+\sqrt{a_{4}-3}=\frac{1}{2} \cdot\left(a_{1}+a_{2}+a_{3}+a_{4}\right)-1
$$

Find the real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ which satisfy the equality

$$
\sqrt{a_{1}}+\sqrt{a_{2}-1}+\sqrt{a_{3}-2}+\sqrt{a_{4}-3}=\frac{1}{2} \cdot\left(a_{1}+a_{2}+a_{3}+a_{4}\right)-1
$$
Point $O$ is located inside triangle $ABC$. Line $BO$ intersects $AC$ at point $E$, and line $AO$ intersects $BC$ at point $D$. Show that $\frac{A O \cdot E C}{A E \cdot O D}=2$, if and only if $B D=D C$.

Solution: The expressions with radicals in the equality make sense if: $a_{1} \geq 0, a_{2} \geq 1, a_{3} \geq 2$ and $a_{4} \geq 3$. We have

$$
\begin{gathered}
\sqrt{1 \cdot a_{1}} \leq \frac{a_{1}+1}{2}, \sqrt{1 \cdot\left(a_{2}-1\right)} \leq \frac{a_{2}-1+1}{2}=\frac{a_{2}}{2}, \sqrt{1 \cdot\left(a_{3}-2\right)} \leq \frac{a_{3}-2+1}{2}=\frac{a_{3}-1}{2} \\
\sqrt{1 \cdot\left(a_{4}-3\right)} \leq \frac{a_{4}-3+1}{2}=\frac{a_{4}-2}{2}
\end{gathered}
$$

Summing up, we get:

$$
\begin{gathered}
\sqrt{a_{1}}+\sqrt{a_{2}-1}+\sqrt{a_{3}-2}+\sqrt{a_{4}-3} \leq \frac{1}{2} \cdot\left(a_{1}+a_{2}+a_{3}+a_{4}\right)+\frac{1}{2}-\frac{1}{2}-1= \\
\frac{1}{2} \cdot\left(a_{1}+a_{2}+a_{3}+a_{4}\right)-1
\end{gathered}
$$

The inequality becomes an equality if the numbers to which we applied the mean inequality are equal. Thus, $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4$.

Answer: $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4$.

8.4 Point $O$ is located inside triangle $ABC$. Line $BO$ intersects $AC$ at point $E$, and line $AO$ intersects $BC$ at point $D$. Show that $\frac{A O \cdot E C}{A E \cdot O D}=2$, if and only if $B D=D C$.