Where Are They Concurrent?

I will use homothety .

A homothety with centre and similarity Coefficient -1 maps $∆ABC to ∆A'B'C'$ . Now let H be the orthocentre. See that BH is perpendicular to AC. $<A'CA =90°$ (angle in a semicircle).

Therefore A'C is perpendicular to AC. Therefore A'C and BH are parallel. Similarly CH and A'B are parallel. Therefore HCA'B is a parallelogram. Since in a parallelogram diagonals bisect each other therefore $HD=DA'$ .

Similarly in other parallelograms $HE=EB'$ and $HF=FE'$ .

Therefore a homothety with Centre H and similarity coefficient $1/2$ maps $∆A'B'C' to ∆DEF$ .

Since H is the centre of dilation, therefore the line joining the corresponding points are concurrent at H, the Orthocentre.

We use vectors.

Set the circumcentre at the origin $O$ . Then $A'= -A$ and $D=\frac{B+C}{2}$ . So the point $H= 2D+(-1)A'= A+B+C$ lies on $DA'$ . Similarly $H$ lies on $EB'$ and $FC'$ . So $H=X$ But $H= A+B+C$ is the vector for the orthocentre of the triangle.

Therefore $X$ is the orthocentre of $\Delta ABC$ .

Image link: http://s12.postimg.org/crymmw065/Untitled.png

Let $H$ be the orthocenter of $\triangle ABC.$ Since $AH \perp BC$ and $B'C \perp BC,$ $AH \parallel B'C.$ Also, since $B'A \perp AB$ and $CH \perp AB,$ $CH \parallel B'A.$ Thus, $AHCB'$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, $H,E,B'$ are collinear. Similarly, $H,D,A'$ and $H,F,C'$ are also collinear. Hence, lines $A'D, B'E, C'F$ are concurrent at the $\fbox{orthocenter}$ of $\triangle ABC.$