If two triangles $ABC$ and $DEF$ are such that the perpendiculars from $A,B,C$ to $EF,FD,DE$ are concurrent, prove that the perpendiculars from $D,E,F$ to $BC,CA,AB$ are concurrent.
I tried to apply Ceva's theorem, but since the endpoints of the cevians are in separate triangles, I got nowhere.
On
If the perpendiculars from $A$ to $EF$, $B$ to $FD$, $C$ to $DF$ concur, then the triangles are othologic and this is property is mutual, meaning that the perpendiculars from $D$ to $BC$, $E$ to $AC$, $F$ to $AE$ concur too. This is called the Maxwell's Theorem.
You can read more on this theorem and the orthologic triangles here.
I don't like the "shaky" proofs given in the link of the first answer. So I am going to suggest another meticulous, careful proof, using only basic methods put together. Plus it shows you a lot of techniques, so it is good for educational purposes (personal opinion).
Let the three concurrent lines through the points $A, B, C$ and perpendicular to the lines $EF, \, FD, \, DE$ respectively intersect in the common point of intersection called $O_{ABC}$. Next, construct the three lines through the point $O_{ABC}$ and perpendicular to the edges $AB, \, BC, \, CA$ respectively and let $Q_{AB}, \, Q_{BC}, \, Q_{CA}$ be their corresponding intersection points, i.e. $Q_{AB}, \, Q_{BC}, \, Q_{CA}$ are the orthogonal projections of the point $O_{ABC}$ onto the edges $AB, \, BC, \, CA$ respectively. Furthermore, draw the tree lines $\sigma_A, \, \sigma_B,\, \sigma_C$ through the points $A, \, B, \, C$ perpendicular to the lines $AO_{ABC}, \, BO_{ABC}, \, CO_{ABC}$ respectively and let the three points $A_1, \, B_1, \, C_1$ be the following intersection points: $$A_1 = \sigma_B \cap \sigma_C, \,\, B_1 = \sigma_C \cap \sigma_A, \,\, C_1 = \sigma_A \cap \sigma_B$$ Consequently, we have constructed the triangle $A_1B_1C_1$ whose edges are parallel to the edges of the triangle $DEF$.
First I am going to prove the statement for the special case of triangles $ABC$ and $A_1B_1C_1$. From there, I am going to derive the general case for $ABC$ and $DEF$.
let $H_{A}, \, H_{B}, \, H_{C}$ be the orthocenters of the triangles $A_1BC, \, AB_1C, \, ABC_1$ respectively and let $P_{BC}, \, P_{CA}, \, P_{AB}$ be the intersection points of the altitudes $A_1H_{A}, \, B_1H_{B}, \, C_1H_{C}$ with the edges $BC, \, CA, \, AB$ respectively. Observe, by construction the lines $A_1H_A, \, \, B_1H_{B}, \, C_1H_{C}$ are perpendicular to the edges $BC, \, CA, \, AB$ respectively and our goal is to show that $A_1H_A, \, \, B_1H_{B}, \, C_1H_{C}$ are concurrent.
Since both $CH_A$ (altitude) and $BO_{ABC}$ are orthogonal to $C_1A_1$ they are parallel to each other. Analogously, both $BH_A$ (altitude) and $CO_{ABC}$ are orthogonal to $A_1B_1$ they are parallel to each other. Consequently, the quad $CH_ABO_{ABC}$ is a parallelogram, hence triangles $BCO_{ABC}$ and $CBH_A$ are congruent and therefore $O_{ABC}Q_{BC}$ and $H_AP_{BC}$ are corresponding (congruent) altitudes, so $CQ_{BC} = BP_{BC}$. Thus, if $M_{BC}$ is the midpoint of edge $BC$ then $Q_{BC}M_{BC} = P_{BC}M_{BC}$.
Let $O$ be the center of the circumcircle of triangle $ABC$. Perform a $180^{\circ}$ rotation around point $O$ (i.e. the central symmetry with center $O$). Denote by $O^*_{ABC}$ the image of $O_{ABC}$ under this rotation. Observe $O_{ABC},\, O, \, O_{ABC}^*$ are collinear and $O_{ABC}O = O_{ABC}^*O.$ Furthermore, the image $h_{BC}$ of line $O_{ABC}Q_{BC}$ under the $180^{\circ}$ rotation around $O$ is parallel to $O_{ABC}Q_{BC}$ and $h_{BC}$ passes through $O^*_{ABC}$, i.e. $h_{BC} \parallel O_{ABC}Q_{BC}, \,\, O_{ABC}^* \,\in\,h_{BC}$, and since $O_{ABC}Q_{BC}$ is perpendicular to $BC$, line $h_{BC}$ is also perpendicular to $BC$.
Let $P_{BC}'$ be the intersection point of $h_{BC}$ and $BC$. Then the quad $P_{BC}'Q_{BC}O_{ABC}O_{ABC}^*$ is a trapezoid and the line $OM_{BC}$ is orthogonal to $BC$ (because $O$ is circumcenter of $ABC$ and $M_{BC}$ is the midpoint of $BC$), meaning it is parallel to both $Q_{BC}O_{ABC}$ and $P'_{BC}O^*_{ABC}$, and on top of that $OM_{BC}$ passes through the midpoint $O$ of edge $O_{ABC}O_{ABC}^*$ of the trapezoid. Therefore, $OM_{BC}$ is midsegment of the trapezoid $P_{BC}'Q_{BC}O_{ABC}O_{ABC}^*$, so its other vertex $M_{BC}$ is the midpoint of $Q_{BC}P_{BC}'$. It follows that $P_{BC}'M_{BC} = Q_{BC}M_{BC} = P_{BC}M_{BC}$ meaning that the two points $P_{BC}$ and $P_{BC}'$ coincide, i.e. $P_{BC} \equiv P_{BC}'$. Thus line $h_{BC} \equiv P_{BC}O_{ABC}^*$ is orthogonal to $BC$ and passes through $P_{BC}$. However, the altitude $A_1H_{A}$ is by construction the line orthogonal to $BC$ and passing through $P_{BC}$. Therefore, line $h_{BC}$ is the line $A_1H_{A}$, i.e. $h_{BC} \equiv A_1H_{A}$ yielding the fact that line $A_1H_A$ is the image of $Q_{BC}O_{ABC}$ under the $180^{\circ}$ rotation around $O$.
Absolutely analogously, one can show that the other two lines $B_1H_B$ and $C_1H_C$ are the images of lines $Q_{CA}O_{ABC}$ and $Q_{AB}O_{ABC}$ respectively under the $180^{\circ}$ rotation around $O$. However, since the three lines $Q_{BC}O_{ABC}, \,\, Q_{CA}O_{ABC}$ and $Q_{AB}O_{ABC}$ intersect at a common point $O_{ABC}$, their three corresponding images $A_1H_A, \,\, B_1H_B$ and $C_1H_C$ also intersect in a common point and that point is $O^*_{ABC}$.
Now we are ready to derive the proof for triangle $DEF$. Since triangles $A_1B_1C_1$ and $DEF$ have parallel corresponding pairs of edges, the two triangles are actually homothetic to each other. Then the three lines $A_1H_A, \, B_1H_B, \, C_1H_C$ which are orthogonal to the edges $BC, \, CA, AB$ respectively, are mapped by the homothety to the three lines through the vertices $D, \, E, \, F$ orthogonal to the edges $BC, \, CA, AB$ respectively. As we have already proved, $A_1H_A, \, B_1H_B, \, C_1H_C$ are concurrent, so their three images are also concurrent.
THE END