25~26高三上·安徽淮北一模·第14题

$\dfrac{3\sqrt{6}}{2}$

【题目】已知点$O$是$\triangle ABC$的外心，直线$AO$与线段$BC$交于点$D$。若$BD=1$，$CD=2$，$AC=3$，则$AB=$_______.

【答案】$\dfrac{3\sqrt{6}}{2}$

【解析】因为点$O$是$\triangle ABC$的外心，所以过$O$作$BC$的垂线，交$BC$于点$E$，则$E$为$BC$的中点.

由题可知，$BC = BD + CD = 3$，所以$CE = BE = \dfrac{3}{2}$，所以$DE = \dfrac{1}{2}$.
因为$AC = 3$，所以$BC = AC$，所以$\triangle ABC$为等腰三角形.

连接$CO$，延长交$AB$于点$F$，则$F$为$AB$的中点.

设$\overrightarrow{CO} = \lambda\overrightarrow{CF}$，$\overrightarrow{AO} = \mu\overrightarrow{AD}$，则$\overrightarrow{OF} = (1-\lambda)\overrightarrow{CF}$，$\overrightarrow{OD} = (1-\mu)\overrightarrow{AD}$.
由$\overrightarrow{CO} + \overrightarrow{OD} = \overrightarrow{CD} = \dfrac{2}{3}\overrightarrow{CB}$，得$\lambda\overrightarrow{CF} + (1-\mu)\overrightarrow{AD} = \dfrac{2}{3}\overrightarrow{CB}$；
所以$\overrightarrow{CB} = \dfrac{3}{2}\lambda\overrightarrow{CF} + \dfrac{3}{2}(1-\mu)\overrightarrow{AD}$.

由$\overrightarrow{CF} + \overrightarrow{FB} = \overrightarrow{CB}$，$\overrightarrow{FB} = \overrightarrow{AF} = \overrightarrow{AO} + \overrightarrow{OF} = \mu\overrightarrow{AD} + (1-\lambda)\overrightarrow{CF}$，
得$(2-\lambda)\overrightarrow{CF} + \mu\overrightarrow{AD} = \overrightarrow{CB}$.

所以
$
\begin{cases}
\dfrac{3}{2}\lambda = 2-\lambda \
\dfrac{3}{2}(1-\mu) = \mu
\end{cases}
$
解得$\lambda = \dfrac{4}{5}$，$\mu = \dfrac{3}{5}$.

设$CF = t$，则$CO = \dfrac{4}{5}t$，$AO = BO = \dfrac{4}{5}t$，所以$AD = \dfrac{5}{3}AO = \dfrac{4}{3}t$，$OD = \dfrac{2}{5}AD = \dfrac{8}{15}t$。

由$CO^2 – CE^2 = OE^2 = OD^2 – DE^2$，得
$
\dfrac{16}{25}t^2 – \dfrac{9}{4} = \dfrac{64}{225}t^2 – \dfrac{1}{4}，
$
所以$\dfrac{80}{225}t^2 = 2$，所以$t^2 = \dfrac{45}{8}$.

所以
$
AB = 2BF = 2\sqrt{BC^2 – CF^2} = 2\sqrt{9 – \dfrac{45}{8}} = \dfrac{3\sqrt{6}}{2}.
$