Triangle orthocenter problem I found a theorem written in a clumsy way. Is this theorem true?
Let $ABC$ be a triangle and $DEF$ triangle made by the base points of altitudes of $ABC$. Then the center of an incircle of $DEF$ is an orthocenter of $ABC$.
And does this holds if $ABC$ is obtuse as I'm not sure how we can define the base points in obtuse triangle case?
 A: Yes. This theorem is completely true. Let $D,E$ and $F$ be the base points of the altitudes passing from $C, B$ and $A$ respectively. Denote the ortho-center of $ABC$ as $H$.  As $\widehat{AFC}=\widehat{BEC}=90$, we conclude $HECF$ is an inscribed quadrilateral. Hence $\widehat{HFE}=\widehat{HCE}$. In the same manner we can prove $\widehat{HFD}=\widehat{HBD}$. But notice as $DECB$ is an inscribed quadrilateral(since $\widehat{BEC}=\widehat{BDC}=90$), so $\widehat{DBE}=\widehat{DCE}$, and we conclude $\widehat{HFE}=\widehat{HFD}$. Denote the base points of the altitudes passing from $H$ and orthogonal to $\overline{DE}, \overline{DF}$ and $\overline{EF}$, $A',B'$ and $C'$ repectively. 
$\widehat{HC'F}=\widehat{HB'F}=90$
$\widehat{HFE}=\widehat{HFD}$
hence we conclude $\widehat{FHC'}=\widehat{FHB'}$
So $HFC'$ is congruent with $HFB'$ as ASA (angle-side-angle), so $\overline{HC'}=\overline{HB'}$
IN the same manner we can prove $\overline{HB'}=\overline{HA'}$, so the theorem is proved.

Now note to the case, when the triangle is obtuse
In this case, the base point lies outside of triangle (indeed it is resulted by drawing lines along each altitude). As we know, and as it is apparent from the bellow picture, each triangle has one incircle, and three excircles. 

In this case the ortho-center of the triangle $ABC$, is the center of one of the excircles of triangle $DEF$. Its proof is completely similar to the previous case, which the triangle $ABC$ was acute. 
