We first consider the situation in which no angle in $ABC$ is obtuse. If $ABC$ is right-angled with hypotenuse $AB$, we have $H_a = H_b = C$, and therefore certainly $H_bH_c = H_cH_a$. Any right-angled triangle $ABC$ therefore certainly has the required property. If $ABC$ is not right-angled, the triangles $AH_bB$ and $AH_aB$ certainly are.

We see that both $H_a$ and $H_b$ lie on the semi-circle with diameter $AB$. If $H_bH_c = H_cH_a$, $H_c$ must be the common point of $AB$ and the bisector of $H_aH_b$, and therefore the mid-point $M_{AB}$ of $AB$. Since $CH_c$ is perpendicular to $AB$, we see that $C$ must lie on the bisector of $AB$, and $ABC$ is therefore isosceles.

Now we assume that one angle in $ABC$ is greater than $90^\circ$.
If we assume $H_bH_c = H_cH_a$, we obtain $|AC| = |BC|$ as before. If, however, we assume $H_aH_c = H_aH_b$, we obtain the situation in the second figure. Because of the right angles between the sides and the altitudes, each of the quadrilaterals $AH_aH_bB$, $CH_bBH_c$ and $AH_aCH_c$ is cyclic. It therefore follows that $\angle H_aH_bC = \angle H_aH_bA = \angle H_aBA = \beta = \angle CBH_c = \angle CH_bH_c$ and $\angle H_bH_cC = \angle H_bBC = 90^\circ - \alpha - \beta = \angle H_aAH_b = \angle H_aAC = \angle H_aH_cC$. We see that $|H_aH_b| = |H_aH_c|$ if and only if $\angle H_aH_bH_c = \angle H_aH_cH_b$, which is equivalent to $2\beta = 2 \cdot (90^\circ - \alpha - \beta)$, or $\alpha + 2\beta = 90^\circ$.

Summing up, we see that exactly the right-angled triangles, the isosceles triangles and triangles in which two angles $\alpha$ and $\beta$ fulfill the equation $\alpha + 2\beta = 90^\circ$ have the required property. $\square$