a.
+ Ta có: $\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC}$
⇔ $(\overrightarrow {AC})^{2}= (\overrightarrow {AB} + \overrightarrow {BC})^{2}$
⇔ $AC^{2} = AB^{2} + BC^{2} + 2AB.AC.cos(\widehat{ABC})$
+ Ta có: $\overrightarrow{BI} = \frac{1}{2} (\overrightarrow{BD} + \overrightarrow{BF})$
⇒ $(\overrightarrow{BI})^{2} = \frac{1}{4} (\overrightarrow{BD} + \overrightarrow{BF})^{2}$
= $\frac{1}{4}(BD^{2} + BF^{2} + 2BD.BF.cos(\widehat{BDF})$
+ Mà: $\left \{ {{AB = AD} \atop {BC = BF}} \right.$ ⇒ cos($\widehat{ABC}$ $=$ cos($\widehat{DBF})$
⇒ $AC^{2} = 4BI^{2}$ ⇒ $AC = 2BI$
b.
+ Ta có: $\left \{ {{\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} } \atop {\overrightarrow{BI} = \frac{1}{2}(\overrightarrow{BC} + \overrightarrow{BF})}} \right.$ ⇒ $\overrightarrow{AC}.\overrightarrow{BI} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow {BC})(\overrightarrow{BD} + \overrightarrow{BF})$ $=$ $\frac{1}{2}(\overrightarrow{AB}.\overrightarrow{BD} + \overrightarrow{BC}.\overrightarrow{BD} + \overrightarrow{AB}.\overrightarrow{BF} + \overrightarrow{BC}. \overrightarrow{BF})$ $=$ $\frac{1}{2}(\overrightarrow{BC}.\overrightarrow{BD} + \overrightarrow{AB}.\overrightarrow{BF})$ $=$ $\frac{1}{2}(BC.BD.cos$$\widehat{DBC} + AB.BF.$ $cos\widehat{ABF})$ $=$ $\frac{1}{2}(BF.BD.cos$$\widehat{DBC} + BF. BD.$ $cos(π - \widehat{ABF}))$ $=$ $\frac{1}{2}(BF.BD.cos$$\widehat{DBC} + BF. BD.(-cos(\widehat{ABF})))$ $(*)$.
+ Mà: $\widehat{DBC} = \widehat{ABF}$ ⇒ $(*) = 0$ ⇒ $AC ⊥ BI$.
