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 !
On
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:
$$ \require{cancel} \begin{align} 2 \overrightarrow{AI} \cdot \overrightarrow{BH} &= (\overrightarrow{AO}+\overrightarrow{AH})\cdot(\overrightarrow{BO}+\overrightarrow{OH}) && \text{by } (4) \\ &= \cancel{\overrightarrow{AO}\cdot\overrightarrow{BO}} + \overrightarrow{AO}\cdot\overrightarrow{OH}+\overrightarrow{AH}\cdot\overrightarrow{BO}+\bcancel{\overrightarrow{AH}\cdot\overrightarrow{OH}} && \text{by } (2), (3) \\ &= (\cancel{\overrightarrow{AH}}+\overrightarrow{HO})\cdot\overrightarrow{OH} + (\bcancel{\overrightarrow{AO}}+\overrightarrow{OH})\cdot\overrightarrow{OC} && \text{by } (2),(3),(1) \\ &= -\overrightarrow{OH}\cdot\overrightarrow{OH} + \overrightarrow{OH}\cdot(\overrightarrow{OH}+\cancel{\overrightarrow{HC}}) && \text{by } (3)\\ &= 0 \end{align} $$
The most elegant solution is the following:
Take $D$ to be the middle of $HC$. Then take a look at $\triangle AOD$.
$DI\parallel OC$ (mid-segment), so $DI\perp AO$.
$OH$ is perpendicular to $AC$ by construction.
So $AI$ is the third altitude: $AI\perp OD$.
But $OD$ is mid-segment of triangle $BHC$ ($OD\parallel BH$)
So $AI\perp BH$, which was to be proved
$\ $
Other proofs (depending on your level):
$\ $
Alternative proof #1: (using similarity)
Consider $E$ the orthogonal projection of $B$ on $AC$ and $F=AI\cap BH$. Then:
$\triangle BEC\sim\triangle AOC\sim\triangle AHO$ (AA case)
So $\frac{CE}{OH}=\frac{CB}{OA}$ and we also have $\frac{CE}{OH}=\frac{CH}{OI}$ (the ratio of halves)
Which means $\triangle OAI\sim\triangle CBH$ (by the SAS similarity case)
Then $\angle OAI=\angle CBH$, so the quadrialater $ABOF$ is inscriptible, so $\angle BFA=\angle BOA=90^{\circ}$
$\ $
Alternative proof #2: (using vectors and the dot product, which is zero iff the vectors are orthogonal, which dxvi already posted in a shorter solution, so you should follow his instead of mine)
$$\require{cancel} \vec{AI}\cdot\vec{BH}=(\vec{AO}+\vec{OI})\cdot(\vec{BC}+\vec{CH})\\ =\cancel{\vec{AO}\cdot\vec{BC}}+\vec{AO}\cdot\vec{CH}+\vec{OI}\cdot\vec{BC}+\cancel{\vec{OI}\cdot\vec{CH}}\\ =\vec{AO}\cdot\vec{CH}+\frac{1}{2}\vec{OH}\cdot2\vec{OC}=\vec{AO}\cdot\vec{CH}+\vec{OH}\cdot\vec{OC}\\ =\vec{AO}\cdot(\vec{CO}+\vec{OH})+\vec{OH}\cdot\vec{OC}=\cancel{\vec{AO}\cdot\vec{CO}}+\vec{AO}\cdot\vec{OH}+\vec{OH}\cdot\vec{OC}\\ =\vec{OH}\cdot(\vec{AO}+\vec{OC})=\vec{OH}\cdot\vec{AC}=0$$
which implies $\vec{AI}\perp \vec{BH}$
$\ $
Alternative proof #3: (using analytic geometry)
Let $A(0,a)$, $B(-b,0)$ and $C(0,b)$
The slope of $AC$ is $-\frac{a}{b}$ and the equation of $AC$ is $y=-\frac{a}{b}x+a$
The slope of $OH$ is $\frac{b}{a}$ (negative reciprocal) and the equation of $OH$ is $y=\frac{b}{a}x$
Then solve the system of the two equations above to calculate the coordinates of $H$, say $(u,v)$ as a function of $a$ and $b$.
Then the coordinates of $I$ are just $\left(\frac{u}{2},\frac{v}{2}\right)$ (midpoint)
Then calculate the slopes of $AI$ and $BH$ and show they are negative reciprocals.