What is the locus of $H$? Given a circle of radius $r$ and a chord $AB$ of length 1 unit, let $C$ be a variable point on the circle; let $I$ be the midpoint of $BC$ and let $H$ be the intersection of the line perpendicular to $AC$ from I. What is the locus of $H$?

 A: Consider $AB$ as the reference axis for angles ($x$ axis), and consider $A$ as the origin. Let $\theta$ be the oriented angle between $AB$ and the (moving) line $AC$.
Let $H_1$ be the (moving) projection of $B$ onto  line $AC$.
Let $D$ be the (fixed) point such that $AD$ is a diameter of the given circle.
Let $\theta_0$ be the (fixed) oriented angle between $AB$ and $AD$.
The locus of $H_1$ is the circle with diameter $AB$ which has the following polar equation: 
$$r=r_1=\cos(\theta).$$
By definition, the locus of $C$ is the initial circle, with polar equation 
$$r=r_2=D_1 \cos(\theta-\theta_0).$$
for a certain fixed $D_1$ (the length of the circles' diameter, $AD$), and a certain fixed $\theta_0$ (which is the angle between $AB$ and $AD$).
$I$, being the midpoint of $H_1$ and $C$ (by Thales' theorem), has the following polar equation:
$$r=(r_1+r_2)/2 \ \ \Longrightarrow \ \ r=(D_1\cos(\theta-\theta_0)+cos (\theta))/2$$
which can be transformed (by expanding $\cos(a-b)=\cos a \sin b + ...$) into an expression: 
$$r=p cos(\theta)+q \sin(\theta)$$
which itself is known to be amenable (by taking $\delta:=\sqrt{p^2+q^2}$) to the form
$$r=\delta cos(\theta-\theta_1)$$
for a certain fixed angle $\theta_1$, which means that the locus is a certain circle passing through $A$ and, as one can verify immediately, passing as well through $B$ (for $\theta=0$). 
A: Given a circle of center $O$, an arc $AB$ fixed, and a variable point $C$ on the circle.
We draw $M$ the midpoint of $BC$, from $M$ draw $MH$ perpendicular to $AC$.
To find: The locus of $H$
Let arc $AB = 2\alpha$ which is constant
Angle $\widehat{ACB}$ = $\alpha$
Draw a circle of diameter $OB$ which passes through $M$ and through $E$ the midpoint of $AB$
Produce $HM$ to cut this circle at $I$
Angle $\widehat{CMH} = 90-\alpha$
Angle $\widehat{IMB} = 90 - \alpha$
Arc $BI$ = 180 - $2 \alpha$ which is constant, so $I$ is fixed
The locus of $H$ is the circle of diameter $AI$.

Proof by Dr.Sami Adham

A: Draw $BG \bot AB$ cutting $HI$ produced at $M$. Then, $AHMB$ is cyclic on the red circle. This circle cuts $BC$ at $N$. 

We let $C’$ be another location of $C$. $AC’$ cuts the red circle at $H’$. $H’N$ is joined. We are going to show that $H’$ is one of the point on the locus of $H$ and that locus is the red circle.
From the fact that $\angle AH’M = 90^0$, we have $MH’ \bot AC’$. We are done if we can prove when $MH’$ cut $BC’$ at $K$ and $BK = KC’$.
By exterior angle of cyclic quadrilateral, $\alpha = \alpha_1 = \alpha_2$. This means $C’C // H’N$.
By angles in the same segment, $\beta = \beta_1 = \beta_2$. This means $KIMB$ is cyclic.
By angles in the same segment, $\gamma = \gamma_1 = \gamma_2$. This means $H’N // KI$.
Applying the intercept theorem to $\triangle BCC’$, we get $BK = KC’$ 
