+ Ta có: $\overrightarrow{AL} = \dfrac {b}{b + c}\overrightarrow{AB} + \dfrac {c}{b + c}\overrightarrow{AC}$.
+ Ta có: $\overrightarrow{CM} = \dfrac {\overrightarrow{CA} + \overrightarrow{CB}}{2} = \dfrac {\overrightarrow{AB} - 2\overrightarrow{AC}}{2}$.
+ Theo giả thuyết:
$AL ⊥ CM ⇔ \overrightarrow{AL}.\overrightarrow{CM} = 0$.
$⇔ (b\overrightarrow{AB} + c\overrightarrow{AC})(\overrightarrow{AB} - 2\overrightarrow{AC}) = 0$.
$⇔ bc^{2} + bc^{2}cosA - 2cb^{2}cosA - 2cb^{2} = 0$.
$⇔ (c - 2b)(1 + cosA) = 0$.
$c = 2b$. (do $cosA > -1$)
+ Khi đó: $CM^{2} = \dfrac {b^{2} + a^{2}}{2} - \dfrac {c^{2}}{4} = \dfrac {a^{2} - b^{2}}{2}$.
$AL^{2} = \dfrac {1}{9}(\overrightarrow{AB} + \overrightarrow{AC})^{2}$.
$= \dfrac {1}{9}(AB^{2} + AC^{2} + 2\overrightarrow{AB}.\overrightarrow{AC}$.
$= \dfrac {2}{9}(9b^{2} - a^{2})$.
+ Ta có: $\dfrac {CM}{AL} = \dfrac {3}{2}\sqrt {5 - 2\sqrt {5}}$.
$⇔ \dfrac {CM^{2}}{AL^{2}} = \dfrac {9}{4}.\dfrac {a^{2} - b^{2}}{9b^{2} - a^{2}} = \dfrac {9}{4}(5 - 2\sqrt {5})$.
$⇔ \dfrac {a^{2} - b^{2}}{9b^{2} - a^{2}} = 5 - 2\sqrt {5}$
$⇔ \dfrac {a^{2}}{b^{2}} = 6 - \sqrt {5}$.
+ Ta có: $cosA = \dfrac {b^{2} + c^{2} - a^{2}}{2bc} = \dfrac {5b^{2} - a^{2}}{4b^{2}} = \dfrac {\sqrt {5} - 1}{4}$.
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