Solution:
Let lines $KH$, $AB$ intersect at $M$ (Figure 5a).
From the quadrilateral $KMAC$ we have
$$\angle KMA = 360^{\circ} - \angle A - \angle ACK - \angle CKM = 360^{\circ} - \angle A - 90^{\circ} - (180^{\circ} - 2\angle KCH) = 90 - \angle A + 2\angle KCH = 90 - \angle A + 2(90^{\circ} - \angle ACB) = 270^{\circ} - \angle A - 2\angle ACB = 270 - \angle A - \angle ACB - \angle ABC = 270^{\circ} - 180^{\circ} = 90^{\circ}.$$

(a)

(b)
Figure 2: Exercise G2.
so $KH \perp AB$ as wanted.

Solution:
Let $D$ be a point on $c$ such that $AD < AC$, and let $E, Z$ be the second points of intersection of lines $AD$ and $BD$ with $c$ respectively. Let also $N$ be the second point of intersection of line $BE$ with the circle $c$. Figure 5b shows $Z$ between $B, D$. The argument below can be trivially modified to apply in case $D$ is in the segment $B, Z$ as well. It is
$$AB^{2} = AC^{2} = AD \cdot AE \Rightarrow \frac{AB}{AE} = \frac{AD}{AB}$$
This relation and the fact that $\angle BAE = \angle BAD$ implies that the triangles $ABE, ADB$ are similar. Thus $\angle ABE = \angle ADB$. Also from the cyclic quadrilateral we get $\angle ADB = \angle ZNE$. Therefore $\angle ABE = \angle ZNE$, so $AB \parallel NZ$.
Call $P$ the intersection point of $BC, NZ$. Since $AC$ is tangent to $c$ it is
$$\angle CEH = \angle BCA$$
and then
$$\begin{aligned} & \angle ZNH + \angle CNZ = \angle HNC = \angle CEH \stackrel{(3)}{=} \angle BCA = \angle ABC = \angle ZPC = \angle BCN + \angle CNZ \\ \Rightarrow \quad & \angle ZNH = \angle BCN \\ \Rightarrow \quad & \angle ZNH = \angle HZN \end{aligned}$$
Therefore $H$ is the midpoint of the arc $NZ$, so $KH \perp NZ$ and as $AB \parallel NZ$ we finally get $KH \perp AB$ as wanted.