1. The problem states that $AH = HB$ and $CH$ is perpendicular to $AB$. We need to analyze the geometric properties given these conditions. 2. Since $AH = HB$, point $H$ is the midpoint of segment $AB$. This means $H$ divides $AB$ into two equal parts. 3. The fact that $CH$ is perpendicular to $AB$ means that $CH$ is the altitude from point $C$ to the line $AB$. 4. Combining these, $CH$ is the altitude drawn from $C$ to the midpoint $H$ of $AB$. This implies that triangle $ABC$ is isosceles with $AC = BC$ because the altitude to the base also bisects the base. 5. Therefore, the key conclusion is that triangle $ABC$ is isosceles with $AC = BC$ due to the altitude $CH$ being perpendicular to $AB$ at its midpoint $H$.