Prove that the altitudes of an acute triangle intersect inside the triangle. 
Prove that the altitudes of an acute triangle intersect inside the triangle.

I can pretty easily see that this is true by a pythagorean theorem argument. Given any two sides, the smaller length to the side will be closer on the base so thus the orthocenter must lie inside the triangle. What would be a more mathematical way of proving this?
 A: By contradiction:

$\triangle ABC$ is acute, $\overline{BD}$ is the altitud of the side $\overline{AC}$, then by definition $m\angle BDC=90°$, and in $\triangle ABC$, $m\angle A+m\angle ABC+m\angle ACB=180°$, in $\triangle CBD$, $m\angle D+m\angle DCB+m\angle CBD=180°\implies m\angle DCB+m\angle CBD<180 $.
By exterior angle $m\angle DCB=m\angle A+m\angle ABC$, now how $\angle ACB$ is acute, then $m\angle ACB<m\angle BDC=90°$, Then
$$m\angle A+m\angle ABC+m\angle ACB<m\angle A+m\angle ABC+m\angle BCD\implies m\angle A+m\angle ABC+m\angle BCD>180$$
then $$m\angle A+m\angle ABC+m\angle BCD=m\angle DCB+m\angle CBD>180$$
this a contradiction
A: Let $\Delta ABC$ be an acute triangle. At vertex $B$ raise a perpendicular on $BC$. Since $\angle ABC$ is acute, $A$ must be on the same side of the perpendicular with $C$. Repeat at $C$, and it follows that $A$ must be in between the two perpendiculars raised at $B, C$.
Now, let $AA'$ be the altitude through $A$. $AA'$ is parallel with the two perpendiculars previously constructed, therefore does not intersect either of them. It follows that all points on the $AA'$ line lie between the two other perpendiculars. In particular, $A'$ must lie between $B$ and $C$, thus the segment $AA'$ is entirely contained inside $\Delta ABC$.
By symmetry, the same applies to the other altitudes, thus to their intersection, the orthocenter.

[EDIT] Followup to the question asked in a comment...

The proof as written above relies on the following proposition. Let $A', B'$ be arbitrary points inside segments $BC, CA$, respectively. Then the intersection $P$ of cevians $AA', BB'$ lies inside the triangle $\Delta ABC$. Depending on background and context, this proposition may be assumed as "obvious", known beforehands, or requiring to be proved.
In the latter case, the proof is a direct consequence of the plane separation postulate: if $M,N$ are points on opposite sides of a line $\ell$ then there exists a point $P$ in segment $MN$ that lies on $\ell$.
Since $B'$ is between $A$ and $C$, points $A$ and $C$ lie on opposite sides of the line $BB'$. By convexity of line segments, point $A'$ on segment $BC$ lies on the same side of $BB'$ as $C$, thus opposite $A$. It follows that segment $AA'$ intersects line $BB'$ at a point $P$ that lies inside the segment $AA'$. By convexity of the triangle, point $P$ must be inside triangle $\Delta ABC$.
Back to the orthocenter case in point, since the altitudes in an acute triangle were proved to be interior cevians, it follows that their intersection is an interior point.
A: Here is a proof
by analytic geometry.
Assume that the vertices are
$A=(0, 0), B=(1, 0)$,
and
$C=(a, 1)$.
It turns out that
the condition for
the intersection of the altitudes
to be inside the triangle
is $0 < a < 1$.
Looking at the triangle,
it is intuitively obvious
that this makes the triangle acute,
and it is nice that
this falls out of the proof.
The vertical altitude
through $C$ is
$x = a$.
The line $BC$
has slope
$\frac{1-0}{a-1}
=\frac{1}{a-1}
$.
Therefore the altitude through $A$
has slope
$1-a$,
so its equation is
$y = (1-a)x$.
Therefore, its intersection with
the altitude through $C$
has
$x=a, y=a(1-a)$,
so the intersection of the altitudes
is it
$V=(a, a(1-a))$.
A point inside the triangle is
$D
=(A+B+C)/3
=(a+1, 1)/3
=((a+1)/3, 1/3)
$.
Therefore $V$ is inside the triangle
if and only if it is
on the same side of each side
as $D$.
The equation of side
$AB$ is
$y=0$.
Evaluating this at $D$
is
$1/3$
and at $V$ is
$a(1-a)$.
Since $1/3 > 0$,
we must have
$a(1-a) > 0$
or
$0 < a < 1$.
The equation of side
$AC$ is
$y - x/a=0$.
Evaluating this at $D$
is
$\frac13-\frac{a+1}{3a}
=\frac{a-(a+1)}{3a}
=\frac{-1}{3a}
$
and at $V$ is
$a(1-a)-a/a
=a-a^2-1
$.
Since $\frac{-1}{3a}$
has the opposite sign of $a$,
and we now know that
$0 < a < 1$,
we must have
$a-a^2-1 < 0$
or
$a(1-a) < 1$.
Since we know that
$0 < a < 1$,
this is true.
The equation of side
$BC$ is
$\frac{1}{a-1}
=\frac{y}{x-1}$
or
$x-1-y(a-1) = 0$.
Evaluating this at $D$
is
$0
=\frac{a+1}{3}-1-\frac13(a-1)
=\frac{a+1-3-(a-1)}{3}
=\frac{-1}{3}
$
and at $V$ is
$x-1-y(a-1)
=a-1-a(a-1)(a-1)
=(a-1)(1-a(a-1))
$.
Since
$\frac{-1}{3} < 0$,
we must have
$(a-1)(1-a(a-1))
< 0$.
Since $0 < a < 1$,
$a-1 < 0$
and
$a(a-1) < 0$
so 
$1-a(a-1) > 1 > 0$,
so we do have
$(a-1)(1-a(a-1))
< 0$.
A: Here is an easy proof which i hope clear enough. I uploaded a picture for easier reference.

First, i hope it's obvious enough that:

Triangle is right if and only if orthocenter is on any segment (more specifically, vertex).

Assume $\triangle ABC$ is an obtuse with $\angle A > 90^\circ$. Draw altitude from $C$. Let point $D$ is an intersection of altitude and $AB$. Since $\triangle ABC$ is obtuse, $[CD$ touches the triangle only at one point which is $C$ itself. Notice that orthocenter must be on $[CD$ and $C$ cannot be orthocenter (otherwise $[BC]$ would be an altitude making triangle non-obtuse); therefore orthocenter must be on outside of triangle. We proved:

If triangle obtuse, orthocenter is outside.

Now, assume $\triangle ABC$ is any triangle (not necessarily obtuse) that does not contain its orthocenter. Draw line $AB$ and altitude from $C$ which intersect the line at $D$. If $D$ is not on segment $[AB]$ then $\triangle ABC$ is obtuse. If $D$ is on segment $[AB]$ then altitude of $A$ intersect $[CD$ inside the triangle (Because $A$ and $B$ are on different sides of $[CD$ and $A$, $D$ and $C$ are colinear) which makes orthocenter inside and contradicting the hypothesis. We proved:

If orthocenter is outside, triangle obtuse.

Therefore, since the right triangles out of the picture:

Triangle acute if and only if orthocenter is inside.

