Prove the lines of the orthocenter are concurrent BY Ceva's theorem Prove that the lines of the orthocenter are concurrent by Ceva's Theorem (or its converse).
Edit: Ceva's theorem is the theorem stating in a triangle $ABC$, if the lines $AX$, $BY$, and $CZ$ ($X$ being on $BC$, and so forth) are concurrent, then:
$(BX/XC) * (CY/YA) * (AZ/ZB) = 1.$
The converse would be that given ^^
 A: Let

*

*the altitude of $A$ intersect line $BC$ at $D$ (i.e. $AD \perp BC$)

*the altitude of $B$ intersect line $CA$ at $E$ (i.e. $BE \perp CA$)

*the altitude of $C$ intersect line $AB$ at $F$ (i.e. $CF \perp AB$)

*$a,b$ and $c$ be the lengths of segments $BC,CA$ and $AB$ respectively

$\triangle ABC$ has no obtuse angle

(source: mathwords.com)
$$\frac{AF}{FB} \frac{BD}{DC} \frac{CE}{EA} = \frac{b \cos A}{a \cos B} \frac{c \cos B}{b \cos C} \frac{a \cos C}{c \cos A} = 1
$$
By Ceva's Theorem, the orthocenter exists.
$\triangle ABC$ has an obtuse angle
Suppose $\angle B > 90^\circ$.  From the diagram below, it can be seen that altitudes $AD$ and $CF$ lie outside $\triangle ABC$.

(source: mathwords.com)
\begin{align}
\frac{AF}{FB} \frac{BD}{DC} \frac{CE}{EA} =& -\frac{b \cos A}{a \cos (180^\circ - B)} \cdot -\frac{c \cos (180^\circ - B)}{b \cos C} \cdot \frac{a \cos C}{c \cos A} \\
=& \frac{b \cos A}{a \cos B} \frac{c \cos B}{b \cos C} \frac{a \cos C}{c \cos A} = 1
\end{align}
By Ceva's Theorem, the orthocenter also exists.
