A point $V$ of the plane containing a triangle $ABC$ is its orthocenter if and only if, at the same time, $AV \perp BC$ and $BV \perp AC$; that is, $AV \cdot BC = 0$ and $BV \cdot AC = 0$. Substituting $BC = BV - CV$ and $AC = AV - CV$, an easy manipulation leads to the equivalent condition in the form of the equality of the scalar products
$$AV \cdot BV = AV \cdot CV = BV \cdot CV. \quad (1)$$
Our goal is thus to find out when the system (1) is satisfied together with the analogous system
$$KV \cdot LV = KV \cdot MV = LV \cdot MV, \quad (2)$$
expressing the fact that the point $V$ is the orthocenter of the triangle $KLM$. We now express the vectors from (2) as linear combinations of the vectors from (1). By hypothesis, there exists a number $p$, $0 < p < 1$, for which
$$AK = pAB, \quad BL = pBC, \quad CM = pCA.$$

Substituting into the first equality $AK = AV - KV$ and $AB = AV - BV$, we get after a small manipulation the first of the following three equalities
$$KV = (1-p)AV + pBV, \ LV = (1-p)BV + pCV, \ MV = (1-p)CV + pAV;$$
the other two can be derived similarly. Taking products, we get
$$KV \cdot LV = (1-p)^2 AV \cdot BV + p(1-p)AV \cdot CV + p(1-p)BV^2 = \\ = (1-p)s + p(1-p)BV^2,$$
where $s$ denotes the common value of the products from (1). Similarly,
$$KV \cdot MV = (1-p)s + p(1-p)AV^2 \quad \text{and} \quad LV \cdot MV = (1-p)s + p(1-p)BV^2.$$
We see that the system (2) is equivalent to the system of equalities
$$p(1-p)AV^2 = p(1-p)BV^2 = p(1-p)CV^2,$$
which, in view of the condition $p(1-p) \neq 0$, is fulfilled if and only if $|AV| = |BV| = |CV|$. The last condition means that the orthocenter $V$ of the triangle $ABC$ coincides with its circumcenter. This happens if and only if the triangle $ABC$ is equilateral.