Orthocenter of triangle $DEF$ is same as the circumcenter of triangle $ABC$ $D,E,F$ are mid points of the sides of the triangle $ABC$,then prove that the orthocenter of triangle DEF is same as the circumcenter of triangle ABC.
I cannnot figure out what coordinates to suppose for A,B,C.I tried taking $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ but calculations go messy and clumsy.Can someone help me in proving this question?
 A: Algebra isn't necessary. Instead, use geometrical definitions: in particular, the circumcentre $O$ of $ABC$ is at the intersection of the perpendicular bisectors of $AB$, $AC$, and $BC$. Do you see how to continue from there?
A: The circumcenter of the triangle $ABC$ is the intersection of the perpendicular bisectors of its sides, i.e. perpendiculars through $D, E, F$. Now, if you join the middle points of the sides you obtain a similar triangle to the original one with its sides parallel to those of the original.
The orthocenter of $DEF$ is the intersection of its altitudes. From the picture you can see that the altitudes of $DEF$ coincide with the perpendicular bisectors of the sides of $ABC$

A: Let's consider a $\triangle ABC$ with side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will be at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$. The mid-points $D, E$ & $F$ of the sides $BC$, $AC$ & $AB$ respectively of $\triangle ABC$ are calculated as follows $$D\equiv\left(\frac{0+a}{2}, \frac{0+0}{2}\right)\equiv\left(\frac{a}{2}, 0\right)$$ 
$$E\equiv\left(\frac{\frac{a\tan C}{\tan B+\tan C}+a}{2}, \frac{\frac{a\tan B\tan C}{\tan B+\tan C}+0}{2}\right)\equiv\left(\frac{a(\tan B+2\tan C)}{2(\tan B+\tan C)},\frac{a\tan B\tan C}{2(\tan B+\tan C)}\right)$$
$$F\equiv\left(\frac{\frac{a\tan C}{\tan B+\tan C}+0}{2}, \frac{\frac{a\tan B\tan C}{\tan B+\tan C}+0}{2}\right)\equiv\left(\frac{a\tan C}{2(\tan B+\tan C)},\frac{a\tan B\tan C}{2(\tan B+\tan C)}\right)$$  Hence, the ortho-center $H$ of $\triangle DEF$ is determined as follows 
Now, the orthocenter H can be calculated by finding intersection point of any two altitude equations. We can get orthocenter as  $$H\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$
& the circumscribed center say $P$ can be calculated as $$P\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$ Hence, we can find that orthocenter (H) of $\triangle ABC$ will coincide with the circumcenter (P) of $\triangle ABC$ i.e. both H & P are same.  
