107. In a convex quadrilateral $A B C D$, given are $|A B|=a,|A D|=b,|B C|=p-a,|D C|=p-b$. Let $O$ be the point of intersection of the diagonals. Denote by $\alpha$ the angle $\widehat{B} \widehat{A C}$. To what does the length $A O$ tend if $\alpha$ tends to zero?

## § 2. Proof Problems
107. Let's first find $\lim _{\alpha \rightarrow 0} \frac{|A O|}{|O C|}$. Denote $\hat{C}=\beta$. We have

$$
\frac{|A O|}{|O C|}=\frac{S_{A B D}}{S_{B D C}}=\frac{\frac{1}{2} a b \sin \alpha}{\frac{1}{2}(p-a)(p-b) \sin \beta}
$$

By the cosine theorem,

$$
\begin{aligned}
& \quad \mid B D_{1^{2}}=a^{2}+b^{2}-2 a b \cos \alpha \\
& \mid B{1^{2}}^{2}=(p-a)^{2}+(p-b)^{2}-2(p-a)(p-b) \cos \beta \\
& a^{2}+b^{2}-2 a b \cos \alpha=(p-a)^{2}+(p-b)^{2}-2(p-a)(p-b) \cos \beta \\
& \cos \beta=\frac{p(p-a-b)+a b \cos \alpha}{(p-a)(p-b)} \\
& \sin \beta=\sqrt{1-\cos ^{2} \beta}=\sqrt{(1-\cos \beta)(1+\cos \beta)}= \\
& \quad=\frac{\sqrt{a b(1-\cos \alpha)\left(2 p^{2}-2 a p-2 b p+a b+a b \cos \alpha\right)}}{(p-a)(p-b)}
\end{aligned}
$$

Considering that as $\alpha \rightarrow 0$

$$
\cos \alpha \rightarrow 1, \frac{\sin \alpha}{\sqrt{1-\cos \alpha}}=\sqrt{2} \cos \frac{\alpha}{2} \rightarrow \sqrt{2}
$$

we get from (1), (2)

$$
\lim _{\alpha \rightarrow 0} \frac{|A O|}{|O C|}=\sqrt{\frac{a b}{(p-a)(p-b)}}
$$

Since $: A C \mid \rightarrow p$, then

$$
\lim _{\alpha \rightarrow 0}|A O|=p \frac{\sqrt{a b}}{\sqrt{a b}+\sqrt{(p-a)(p-b)}}
$$