<s>[INST] Prove that $\vec{AI}.\vec{BH}=0$ First I'm sorry I don't know how to draw the triangle on LaTeX.
We consider $\triangle ABC$ isoscele in $A$. We denote $O$ the midpoint of $[BC]$ and $H$ the orthogonal projection of $O$ on $(AC)$. We also note $I$ the midpoint of $[OH]$.
Is it a good idea to use Apollonius theorem in $\triangle AOH$ and $\triangle HBC$ ?
In that case we will have : 
$AO^2+AH^2=\frac{1}{2}OH^2+2AI^2$ and $HB^2+HC^2=\frac{1}{2}BC^2+2OH^2$ and maybe find an expression without $OH^2$ will help.
Thanks in advance !
 [/INST] Since the title invites a vector proof, note that by construction:
$$
\begin{align}
& \overrightarrow{BO}=\overrightarrow{OC} && \text{since } O \text{ is the midpoint of } BC \tag{1} \\
& \overrightarrow{AO} \cdot \overrightarrow{BC}=0 && \text{since } \triangle ABC \text{ is isosceles, so } AO \perp BC\tag{2} \\
& \overrightarrow{OH} \cdot \overrightarrow{AC}=0 && \text{since } OH \perp AC \tag{3} \\
& \overrightarrow{AI} = \frac{1}{2}(\overrightarrow{AO}+\overrightarrow{AH}) && \text{since }  I \text{ is the midpoint of } OH \tag{4} 
\end{align}
$$
Then:
$$
\require{cancel}
\begin{align}
2 \overrightarrow{AI} \cdot \overrightarrow{BH} &= (\overrightarrow{AO}+\overrightarrow{AH})\cdot(\overrightarrow{BO}+\overrightarrow{OH}) && \text{by } (4) \\
 &= \cancel{\overrightarrow{AO}\cdot\overrightarrow{BO}} + \overrightarrow{AO}\cdot\overrightarrow{OH}+\overrightarrow{AH}\cdot\overrightarrow{BO}+\bcancel{\overrightarrow{AH}\cdot\overrightarrow{OH}} && \text{by } (2), (3) \\
 &= (\cancel{\overrightarrow{AH}}+\overrightarrow{HO})\cdot\overrightarrow{OH} + (\bcancel{\overrightarrow{AO}}+\overrightarrow{OH})\cdot\overrightarrow{OC} && \text{by } (2),(3),(1) \\
 &= -\overrightarrow{OH}\cdot\overrightarrow{OH} + \overrightarrow{OH}\cdot(\overrightarrow{OH}+\cancel{\overrightarrow{HC}}) && \text{by } (3)\\
 &= 0
\end{align}
$$
</s>